A man weighs 70 kg. He stands on a weighing scale in a lift which is moving:
(a) upwards with uniform speed of 10m/s,
(b) downwards with uniform acceleration of 5 m/s2 and
(c) upwards with uniform acceleration of 5 m/s2.
What would be the reading on the scale in each case?
(d) What would be the reading, if the lift mechanism failed and it hurtled down freely under gravity?
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Retarding Force, F = - 50 N
Mass of the body, m = 20 kg
Initial velocity of the body, u = 15 m/s
Final velocity of the body, v = 0
Acceleration produced in the body can be calculated as:
F = ma
-50 = 20 x a
a = 
Using the first equation of motion,
v = u + at
Time (t) taken by the body to come to rest can be calculated as,
Therefore,
t = 
A pebble of mass 0.05 kg is thrown vertically upwards. Give the direction and magnitude of the net force on the pebble,
(a) during its upward motion,
(b) during its downward motion,
(c) at the highest point where it is momentarily at rest. Do your answers change if the pebble was thrown at an angle of 45° with the horizontal direction?
Ignore air resistance.
i) T is the correct answer.
A particle connected to a string is revolving in a circular orbit around the center.
For rotation, the centripetal force is provided by the tension produced in the string.
Therefore, the net force produced on the particle is the tension, T.
That is, 
where,
F is the net force acting on the particle.
Give the magnitude and direction of the net force acting on a stone of mass 0.1 kg,
(a) just after it is dropped from the window of a stationary train,
(b) just after it is dropped from the window of a train running at a constant velocity of 36 km/h,
(c) just after it is dropped from the window of a train accelerating with 1 m s-2,
(d) lying on the floor of a train which is accelerating with 1 m s-2, the stone being at rest relative to the train.
Neglect air resistance throughout.
a) We have here,
Mass, m = 0.1 kg
Acceleration due to gravity, a = +g = 10 m/s2
Net force, F = ma = 0.1 × 10 = 1.0 N, acting in vertically downward direction.
b) When the train is running at a constant velocity, its acceleration = 0.
Due to this motion, no force is acting on the stone.
Therefore,
Force on the stone, F = weight of stone = mg
= 0.1 × 10 = 1.0 N
This force acts in the vertically downward direction.
c) Acceleration of the train, a = 1 m s-2
Additional force, F' = ma = 0.1 × 1 = 0.1 N, acts on the stone in the horizontal direction.
But once the stone is dropped from the train, F' becomes zero.
Net force on the stone, F = mg = 0.1 × 10
= 1.0 N, acting vertically downwards.
d) When the stone is lying on the train, its acceleration is same as that of the train.
Therefore, the force acting on stone, F = ma
= 0.1 × 1 = 0.1 N
This force is along the horizontal direction of motion of the train.
In each case, the weight of the stone is being balanced by the normal reaction.
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