Force and Motion
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Chapter – 7
Force and motion
Force is an external agent which changes the state, direction and shape and size of a body. Its SI unit is Newton ‘N’ and CGS unit is Dyne.
Gravitational force
The force of attraction between two heavenly bodies towards their centre is called gravitational force. Due to the application of gravitational force, the universe is sustaining. Its SI unit is Newton and CGS is dyne.
Newton’s Law of Gravitation (1687)
This law was introduced by Sir Isaac Newton, a British scientist in 1687 A.D. which is called Newton’s law of gravitation.
Let us consider two heavenly bodies of masses m1 and m2 separated by distance d between their centres.
According to the Newton’s law of gravitation, gravitational force is directly proportional to product of their masses i.e.
Inversely proportion to square of the distance between their centres.
Combining eqn (1) and (2),
F = Gm1m2/d²
Where G is called universal gravitational constant having value 6.67 × 10−11 Nm²/kg² (S.I.) or 6.67 × 10−8 dyne cm²/g² (c.g.s).
G is called universal gravitational constant because its value is same anywhere or everywhere around the universe.
Universal Gravitational Constant (G)
When, m1 = 1 kg, m2 = 1 kg, d = 1 m, then F = G.
The force of attraction between two heavenly bodies of unit masses separated by unit distance is called universal gravitational constant.
Cases:
(i) When one of the mass is doubled and another mass tripled and distance is quartered.
1st condition:
F1 = Gm1m2/d² …. (1)
2nd condition:
m1 = 2m1
m2 = 3m2
d = d/4
Then,
F2 = G(2m1)(3m2)/(d/4)²
F2 = 6Gm1m2/(d²/16)
F2 = 96Gm1m2/d²
F2 = 96F1 (∵ F1 = Gm1m2/d²)
When one of the mass is doubled and other is tripled and distance is quartered, the gravitational force becomes 96 times to the initial force.
(ii) When one of the mass is tripled and one is at unit place and distance is half.
1st condition:
F1 = Gm1m2/d² …. (1)
2nd condition:
m1 = 3m1
m2 = m2
d = d/2
Then,
F2 = G(3m1)(m2)/(d/2)²
F2 = 3Gm1m2/(d²/4)
F2 = 12Gm1m2/d²
F2 = 12F1 (from 1st condition).
When one of the mass is tripled and one is at unit place and distance is half, the gravitational force becomes 12 times to the initial force.
(iii) When both the masses are doubled and distance is halfed.
1st condition:
F1 = Gm1m2/d² …. (1)
2nd condition:
m1 = 2m1
m2 = 2m2
d = d/2
Then,
F2 = G(2m1)(2m2)/(d/2)²
F2 = 4Gm1m2/(d²/4)
F2 = 16Gm1m2/d²
F2 = 16F1 (from condition 1).
When both of the masses are doubled and distance is halfed, the gravitational force becomes 16 times to the initial force.
(iv) When one of the object’s mass is in unit place and another is doubled and distance is at unit place.
1st condition:
F1 = Gm1m2/d²
2nd condition:
m1 = m1
m2 = 2m2
d = d
Then, F2 = Gm1(2m2)/d²
F2 = 2Gm1m2/d²
F2 = 2 × F1 (from condition 1st).
When one of the mass is at unit place and another is doubled and distance is at unit place, the gravitational force becomes 2 times to the initial force.
(v) When both masses are tripled and distance is quarter.
1st condition:
F1 = Gm1m2/d²
2nd condition:
m1 = 3m1
m2 = 3m2
d = d/4
Then, F2 = G(3m1)(3m2)/(d/4)²
F2 = 9Gm1m2/(d²/16)
F2 = 144Gm1m2/d²
F2 = 144 × F1 (from condition 1st).
When both masses are tripled and distance is quarter, the gravitational force becomes 144 times to the initial force.
(vi) When both masses are at unit place and distance is halfed.
1st condition:
F1 = Gm1m2/d²
2nd condition:
m1 = m1
m2 = m2
d = d/2
Then, F2 = Gm1m2/(d/2)²
F2 = Gm1m2/(d²/4)
F2 = 4Gm1m2/d²
F2 = 4F1.
When both masses are at unit place and distance is halfed, the gravitational force becomes 4 times to the initial force.
Acceleration due to Gravity (g)
The acceleration produced in falling object by the action of gravity of the heavenly body is called acceleration due to gravity (g). It is given by:
Where, G = Gravitational constant, M = mass of the heavenly body, R = radius of heavenly body.
If an object is at a height (h) or its depth (h) from the surface of heavenly body, then acceleration due to gravity becomes:
At some height or depth (h) the acceleration due to gravity becomes:
- At equator region of earth (g) = 9.79 m/s²
- At polar region of earth (g) = 9.83 m/s²
- At surface of Jupiter (g) = 25.83 m/s²
- At surface of moon (g) = 1.63 m/s²
Proof:
Let us consider a heavenly body of mass (M), radius (R). Let us consider an object of mass (m) at the surface of the heavenly body.
Now, from Newton’s law of Gravitation, gravitational force is given by:
Again, from Newton’s second law of motion:
From eqn (i) and (ii):
g = GM/R²
If an object is at a height of ‘h’ from surface of heavenly body then acceleration due to gravity is given by:
Gravity at center of earth is 0 m/s².
Coin and feather Experiment
The experiment concludes if there is presence of external resistance i.e. air resistance, the coin falls faster because the air resistance reduces the acceleration due to gravity of feather.
If there is no presence of external resistance i.e. air resistance then both coin and feather falls at same rate because in vacuum acceleration due to gravity is independent to the mass of falling object.
Freefall & Weightlessness
When an object falls freely with the by the action of gravity neglecting external resistance is known as free fall.
During freefall acceleration due to gravity (g) = acceleration due to gravity (a).
Now, Net weight of falling object:
= mg − mg
= 0
i.e. weightlessness
During freefall a body falls only by the action of gravity and weight of falling object now becomes zero is called weightlessness.
Equation of Motion
The equation which give relation between velocity, time, distance, acceleration etc are called equation of motion.
(i) For upward motion:
v² = u² + 2as
s = ut + ½gt²
(ii) For upward motion:
a = −g
v = u − gt
v² = u² − 2gh
h = ut + ½(−g)t²
Conversion of G
G = 6.67 × 10−11 Nm²/kg²
= 6.67 × 10−11 × 105 dyne × 1000 cm × 100 cm / (1000 gm × 1000 gm)
= 6.67 × 10−11 × 105+4 dyne cm² / 106 gm²
= 6.67 × 10−8 dyne cm²/gm².
(iii) For down ward motion
| (a) For thrown object | (b) For dropped object |
|---|---|
| u = constant a = g v = u + gt v² = u² + 2gs h = ut + ½gt² | u = 0 v = 0 + gt v² = 2gh h = ½gt² |
Exercise
1. Choose the correct option for the following questions!
(a) What is the relation between the distance between two object (d) and the gravitational force (F) produced between them?
→ Ans: (iii) F ∝ 1/d²
(b) What is the change in the gravitational force between two objects when their mass is doubled?
→ Ans: (ii) the force becomes four times.
(c) If the gravitational force between two objects on Earth is 60 N, what is the gravitational force between those two objects on the moon?
→ Ans: (i) 10 N.
(d) Which one of the following statements is correct?
→ Ans: (ii) The value of acceleration due to gravity decreases as the height above the surface of the Earth decreases.
(e) At which of the following places you weigh the most?
→ Ans: (iii) Kathmandu or Jhapa.
(f) The radius of the earth is 6371 km and the weight of an object on the earth is 800 N. What is the weight of the object at a height of 6371 km from the surface of the earth?
Given, Radius of Earth (R) = 6371 km
Weight on surface (Ws) = 800 N
Height above surface (h) = 6371 km
Weight on height (Wh) = ?
Weight on height: Wh = GMm/(R+h)² …. (i)
Weight on surface: Ws = GMm/R² …. (ii)
Taking the ratio of (i) and (ii): Wh/Ws = (R/(R+h))²
Wh = Ws(R/(R+h))²
Putting values,
Wh = 800 (6371/(6371+6371))²
Wh = 800 (1/2)²
Wh = 800 × 1/4
∴ Wh = 200 N
Ans: (iii) 200 N.
(g) If the mass and the radius of the celestial body are two times the mass and the radius of earth respectively, what is the value of acceleration due to gravity of that body?
→ Ans: 4.9 m/s².
(h) What will be the weight of the same man on the moon if his weight on earth is 750 N? (The acceleration due to the gravity of moon = 1.63 m/s²).
Given, Weight on earth (We) = 750 N
gravity of earth (ge) = 9.8 m/s²
gravity of moon (gm) = 1.63 m/s²
m = We/ge = 750/9.8 = 76.53 kg
Weight on moon (Wm) = m × gm
= 76.53 × 1.63
= 124.79 N
Ans: 124.79 N.
(i) The mass of planet B is twice the mass of planet A but the radius is half of the radius of planet A. Similarly, the mass of planet C is half of the mass of planet A, but its radius is twice the radius of planet A. If the weight of an object in planets A, B and C is W1, W2 and W3 respectively, which of the following order is correct?
Ans: W2 > W1 > W3.
(j) Which one of the following conclusions is correct while assuming a freely falling object every second?
→ Ans: Velocity increases uniformly.
2. Differentiate between:
(a) Gravitational constant G and acceleration due to gravity g
| Gravitational constant G | Acceleration due to gravity g |
|---|---|
| The force of attraction between two heavenly bodies of unit masses and separated by unit distance is called Gravitational constant. | The acceleration produced in freely falling object due to the force of gravity is called acceleration due to gravity (g). |
| It is denoted by ‘G’. | It is denoted by ‘g’. |
| Its value remains same (6.67 × 10−11) everywhere and there is no change around the universe. | Its value depends upon mass and radius of planet or heavenly body. |
| SI unit is Nm²/kg². | SI unit is m/s². |
(b) Mass and Weight
| Mass | Weight |
|---|---|
| The total quantity of matter present in an object is its mass. | Weight is the measure of the force of gravity acting on an object. |
| It is scalar quantity. | It is vector quantity. |
| It is always same everywhere in the universe. | It depends upon mass and acceleration due to gravity. |
| Its SI unit is kilogram (kg). | Its SI unit is newton (N). |
4. Answer the following questions:
(a) What is gravity?
→ The force exerted by the planet or satellite pulling their nearby objects towards their center is called gravity or the force of gravity.
(b) State Newton’s universal law of gravitation.
According to the Newton’s law of gravitation, gravitational force is directly proportional to the product of masses:
Inversely proportional to the square of the distance between their centres:
Solving eqn (i) and (ii):
F = Gm1m2/d² (G is a constant)
Where G is the universal gravitational constant having the value 6.67 × 10−11.
(c) What is gravitational force?
→ The mutual gravitational force is attraction force between any 2 objects towards their center.
(d) Define gravitational constant (G).
→ The force of attraction between two heavenly bodies of unit masses and separated by unit distance is called gravitational constant.
(e) Under what conditions is the value of gravitational force equal to the gravitational constant (F = G)?
→ When both mass are at unit place and distance is at unit place, then the value of gravitational force will be equal to the gravitational constant.
When, m1 = 1 kg
m2 = 1 kg
d = 1 m
Putting the value in F = Gm1m2/d²
F = G × 1 × 1/(1)²
F = G.
(f) Write two effects of gravitational force.
→ Following are the two effects of gravitational force:
- Earth revolving around sun.
- Moon revolve around sun.
(g) Mathematically present the difference in the gravitational force between two objects when the mass of each body is made doubled and distance between them is made one half their initial distance.
1st case:
F1 = Gm1m2/d²
2nd case:
m1 = 2m1, m2 = 2m2, d = d/2
F2 = G(2m1)(2m2)/(d/2)²
F2 = 4Gm1m2/(d²/4)
F2 = 16Gm1m2/d²
F2 = 16F1.
The gravitational force will be 16 times greater than the initial gravitational force.
(h) What is gravitational force?
→ The force exerted by the planet or satellite on nearby object is often called the force of gravity. It is an invisible pull that exists between objects with mass.
(i) Define acceleration due to gravity.
→ The acceleration produced on a freely falling object due to gravity is called acceleration due to gravity.
(j) What is free fall? Give two examples of it.
→ When an object falls freely only by the action of gravity neglecting external resistance is known as free fall. Its examples are given below:
- Landing with a parachute in moon.
- Dropping a coin.
(k) Under what conditions is an object said to be in free fall?
→ An object is said to be in free fall when it is only influenced by gravity and no other forces such as air resistance or external forces are acting upon it.
(l) Write the conclusions of the feather and coin experiment.
→ The conclusions of coin and feather experiment are:
- In the absence of air resistance, objects with different masses like a feather and a coin fall at the same rate.
- Gravity affects all objects equally regardless of their mass when there is no air resistance.
(m) What is weightlessness?
→ During freefall a body falls only by the action of gravity and weight of falling object now becomes zero is called weightlessness.
(n) Mention any four effects of gravitational force.
→ Following are the four effects of gravitational force:
- Object thrown upward falls down on the ground.
- Existence of universe and solar system.
- Earth revolving around sun.
- Tides in the ocean due to gravitational force between earth and moon.
(o) Prove that acceleration due to the gravity of the earth is inversely proportional to the square of its radius (g ∝ 1/R²).
Let us consider a heavenly body of mass (M), radius (R), and an body of mass (m) on the body of the planet at surface.
Now, from Newton’s law of universal gravitational force is given by:
Again, from Newton’s second law of motion:
From equations (i) and (ii):
g = GM/R² …. (iii)
Since the mass of falling object is not included in the equation and in eqn (iii) both G and M are constants,
The acceleration due to gravity is inversely proportional to the square of the radius of the heavenly body.
(p) Mention the factors that influence acceleration due to gravity.
→ Following are the factors that influence acceleration due to gravity:
- height from the surface.
- radius of heavenly body.
- mass of heavenly body.
(q) The acceleration due to gravity in the Earth’s surface is 9.8 m/s². What does this mean?
→ The acceleration due to gravity in the Earth’s surface is 9.8 m/s² means that the falling object falls 9.8 m/s in each second.
OR
It means that every second an object near the earth’s surface experiences an increase in velocity by 9.8 m/s² due to the force of gravity.
(r) Mass of the moon is about 1/81 times the mass of the Earth’s and its radius is about 37/10 times the radius of the earth. If the earth is squeezed to the size of the moon, what will be the effect on its acceleration due to gravity? Explain with the help of mathematical calculation.
Given, mass of moon (Mm) = 1/81
mass of earth (Me) = Me/81
Radius of moon (Rm) = 37/10 Re
Now, g = GMe/Re²
= GMe/(37/10 Re)²
= 100/1369 × GMe/Re²
= 100/1369 × ge (∵ ge = GMe/Re²)
= 100/1369 × 9.8
= 0.71 m/s².
(s) The acceleration due to gravity of an object of mass 1 kg in outer space is 2 m/s². What is the acceleration due to gravity of another object of mass 10 kg at the same point? Justify with arguments.
→ The acceleration due to gravity of another object of mass 10 kg at same point is same because acceleration doesn’t depends upon mass of an object.
(t) A mountaineer measures the mass and weight of an object in the mountain and then in the Terai. Compare the data he obtains.
→ When measuring the mass and weight of an object in the mountain and Terai region, the mass of object will remain the same in both place. However, the weight of the object will depend on the earth due to gravity which can be different at different location on Earth, i.e. w = mg.
(u) A student suggests a trick for gaining profit in a business. He suggests buying oranges from the mountain, selling them to Terai at the cost price. If a beam balance is used during this transaction, explain, based on scientific fact, whether his trick goes wrong or right?
→ The student’s trick will go wrong and lead to no profit. This is because he is using a beam balance which measures the mass of the object which doesn’t varies upon the place. If he used a spring balance which measures the weight, then he can get profit.
(v) How is it possible to have a safe landing while jumping from a flying airplane using a parachute? It is possible to have a safe landing on the moon in the same way? Explain with reasons.
→ Because the parachute increases the air resistance which reduces the force of gravity and slow down the speed while falling. However, on the moon, there is no atmosphere to provide air resistance so a parachute alone will not work for a safe landing on the moon like as on Earth.
(w) The acceleration of an object moving on the earth is inversely proportional to the mass of the object, but for an object falling towards the surface of the earth, the acceleration doesn’t depend on the mass of the object. Why?
→ This is because gravity affects all objects equally during free fall, causing them to accelerate at the same rate, regardless of their mass. In this scenario, the mass doesn’t factor into the equation, resulting in the same acceleration for all objects.
5. Numericals:
(a) The masses of two objects A and B are 20 kg and 40 kg respectively. If the distance between their centers is 5 m, calculate the gravitational force produced between them.
Given, m1 = 20 kg
m2 = 40 kg
d = 5 m
We K.T.,
F = Gm1m2/d²
F = 6.67 × 10−11 × 20 × 40/(5)²
F = 53.36 × 10−9/25
F = 2.1344 × 10−9 N.
(b) Mass of the sun and Jupiter are 2 × 1030 kg and 1.9 × 1027 kg respectively. If distance between sun and Jupiter is 1.8 × 108 km, calculate the gravitational force between sun and Jupiter.
Given, mass of sun (m1) = 2 × 1030 kg
mass of Jupiter (m2) = 1.9 × 1027 kg
distance (d) = 1.8 × 108 km = 1.8 × 1011 m
F = Gm1m2/d²
= 6.67 × 10−11 × 2 × 1030 × 1.9 × 1027 / (1.8 × 1011)²
= 25.346 × 1046 / (3.24 × 1022)
= 25.346 × 1046 / 3.24 × 1022
= 7.82 × 1024.
(c) Gravitational force produced between the Earth and the Sun is 3.54 × 1022 N. If the masses of the Earth and Sun are 5.972 × 1024 kg and 2 × 1030 kg respectively, what is the distance between them?
Gravitational force (F) = 3.54 × 1022 N
m1 = 5.972 × 1024 kg
m2 = 2 × 1030 kg
d = ?
F = Gm1m2/d²
d² = Gm1m2/F
d = √(Gm1m2/F)
d = √[(6.67 × 10−11 × 5.972 × 1024 × 2 × 1030)/(3.54 × 1022)]
d = √(22.5048 × 1021)
d = √(2.25048 × 1022)
d = 1.500 × 1011 m.
(d) Gravitational force between Earth and Moon is 2.01 × 1020 N and distance between their centres is 3.84 × 105 km and the mass of the earth is 5.972 × 1024 kg. Calculate the mass of the moon.
Given, F = 2.01 × 1020 N
d = 3.84 × 105 km = 3.84 × 108 m
m1 = 5.972 × 1024 kg
m2 = ?
F = Gm1m2/d²
2.01 × 1020 = 6.67 × 10−11 × 5.972 × 1024 × m2 / (3.84 × 108)²
m2 = 7.39 × 1022 kg.
(e) The mass of the moon is 7.302 × 1022 kg. If the average distance between the earth and the moon is 384400 km, calculate the gravitational force exerted by the moon on 1 kg mass of water at the surface of the earth.
Given, m1 = 7.302 × 1022 kg
m2 = 1 kg
d = 384400 km = 3.844 × 108 m
F = Gm1m2/d²
= 6.67 × 10−11 × 7.302 × 1022 × 1 / (3.844 × 108)²
= 3.31 × 10−5 N.
(g) If the mass of the moon is 7.332 × 1022 kg and its radius is 1737 km, calculate its acceleration due to gravity.
Given, mass of moon = 7.332 × 1022 kg
Radius (R) = 1737 km = 1.737 × 106 m
g = Gm/R²
= 6.67 × 10−11 × 7.332 × 1022 / (1.737 × 106)²
= 1.63 m/s².
(h) Mass of the earth is 5.972 × 1024 kg and the diameter of the moon is 3472 km. If the earth is compressed to the size of the moon, how many times will be the change in acceleration due to gravity of the earth so formed than that of the real earth?
Given, m = 5.972 × 1024 kg
d = 3472 km
R = 3472/2 = 1736000 m = 1.736 × 106 m
g = Gm/R²
= 6.67 × 10−11 × 5.972 × 1024 / (1.736 × 106)²
= 132.0824 m/s².
comparing, 132.0824/9.8 = 13.48 times.
(i) If the mass of Mars is 6.9 × 1023 kg and its radius is 3389 km, calculate the weight of an object of mass 200 kg on the surface of Mars.
Given, Mmars = 6.9 × 1023 kg
Rmars = 3389 km = 3.389 × 106 m
g = GMmars/(Rmars)²
= 6.67 × 10−11 × 6.9 × 1023 /(3.389 × 106)²
= 3.716 m/s².
Now, W = 200 × 3.716 = 750 N.
(j) The acceleration due to the gravity of the earth is 9.8 m/s². If the mass of Jupiter is 319 times the mass of the Earth and its radius is 11 times the radius of the Earth, calculate the acceleration of gravity of Jupiter. What is the weight of an object of mass 100 kg on Jupiter?
(g) If the mass of the moon is 7.392 × 1022 kg and its radius is 1737 km, calculate its acceleration due to gravity.
Given, mass of moon = 7.392 × 1022 kg
Radius (R) = 1737 km = 1.737 × 106 m
g = Gm/R²
= 6.67 × 10−11 × 7.342 × 1022/(1.737 × 106)²
= 1.63 m/s².
(h) Mass of the earth is 5.972 × 1024 kg and the diameter of the moon is 3472 km. If the earth is compressed to the size of the moon, how many times will be the change in acceleration due to gravity of the earth so formed than that of the real Earth?
Given, m = 5.972 × 1024 kg
h = 98 km = 98 × 103 m
R = 6371 × 1000 m = 6.371 × 106 m
m = 10 kg
g = Gm/(R+h)²
= 6.67 × 10−11 × 5.972 × 1024 / (6.371 × 106 + 98 × 103)²
= 9.78 m/s².
Now, W = mg = 10 × 9.78 = 98 N.
(m) The acceleration due to gravity of the Mars is 3.75 m/s². How much mass can a person lift on Mars who can lift 100 kg mass on the Earth?
Given, gravity on Mars = 3.75 m/s²
m = 100 kg
g = 9.8 m/s²
W = m × g = 100 × 9.8 = 980 N.
On Mars:
g = 3.75 m/s²
W = 980 N
W = m × g
980 = m × 3.75
m = 980/3.75
∴ m = 261.33 kg.
(n) When a stone is dropped from a bridge over a river into the water, after 2.5 seconds the sound of the stone hitting the surface of the water is heard. Calculate the height of the bridge from the surface of the water.
Soln, Time (t) = 2.5 sec
Height (h) = ?
gravity (g) = 9.8 m/s²
We K.T., h = ut + ½gt²
h = 0 × 2.5 + ½ × 9.8 × (2.5)²
h = 4.9 × 6.25
∴ h = 30.625 m.
(o) If a stone is dropped from a bridge over a river into the water, after 2.5 seconds the sound of the object. Find the velocity.
Given, u = 0 m/s
height (h) = 15 m
time (t) = ?
We K.T., h = ut + ½gt²
15 = 0 + ½ × 9.8 × t²
15 = 4.9t²
15 = 4.9t²
t² = 3.06
t = √3.06
∴ t = 1.75 s.
v = u + gt
= 0 + 9.8 × 1.75
= 17.19 m/s².
(p) If a cricket ball is thrown vertically upwards into the sky with a velocity of 15 m/s, to what maximum height will the ball reach?
Given, v = 15 m/s
u = 0 m/s
g = 9.8 m/s
h = ?
We know that, h = u + ½gt²
15 = 0 × t + ½ × 9.8 × t²
4.9t² = 15
t² = 3.06
t = √3.06
∴ t = 1.75 s.
Again, v = u + gt
= 0 + 9.8 × 1.75
= 17.19 m/s.
3. Give reasons:
(a) Acceleration due to gravity is not the same in all parts of the earth.
→ Acceleration due to gravity is not the same in all places of earth because it is not constant at a place and location due to gravity depends from the distance from the center of earth.
g ∝ 1/R² F ∝ g
(b) Jumping from a significant height may cause more injury.
→ Jumping from a significant height may cause more injury as the velocity gained during the fall increase with height.
(c) Mass of Jupiter is about 319 times the mass of the Earth but its acceleration due to gravity is only about 2.5 times the acceleration due to gravity of the Earth.
→ Because the acceleration due to gravity not only dependent on the mass of an object but also its radius.
(d) When an object is dropped from same height in the polar region and the equatorial region of the earth, object dropped in the polar region falls faster.
→ Because in polar region it is closer to the center of the earth and its gravitational acceleration is greater.
g ∝ 1/r²
so that the object falls faster in polar region.
(e) Out of two paper sheets, one is folded to form a ball and the paper ball and sheet of paper are dropped simultaneously from same height; paper ball falls first.
→ Because the surface area of folded paper is less compared to unfolded paper and air resistance is also a main form factor for falling object.
(f) When a coin and a feather are dropped simultaneously in a vacuum, they reach at the ground together.
→ Because in vacuum the mass and the feather experience the same acceleration due to gravity since air resistance eliminated.
(g) As you climb Mount Everest, the weight of the goods that you carry decreases.
→ Because the acceleration due to gravity decreases with increase in a distance from the center of the earth such that:
∴ weight ↓
(h) It is difficult to lift a big stone on the surface of the earth but it is easy to lift in water.
→ Because the big stone has more mass and than smaller stone has smaller mass and as we know weight of an object is proportional to mass of object which is F = mg.
(i) Mass of an object remains same but its weight varies from place to place.
→ Because mass is same anywhere in universe and weight is the force exerted on an object due to gravity and it depend upon height and gravity.
(j) One will have eerie feeling when he/she moves down while playing a Rote Ping.
→ Because due to sudden change of height and action in rote ping the person will be in free fall and feel weightlessness and weight becomes zero.
Discussion
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