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Laws of Motion

Question
CBSEENPH11020060
Wired Faculty App

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.

Solution

Irrespective of the direction of motion, acceleration due to gravity always acts downward.
The gravitational force is the only force that acts on the pebble in all three cases.
Its magnitude is given by Newton’s second law of motion as:
                       F = m x a
where,
 
F = net force, 
m= mass of the pebble, and 
a is the acceleration.
Here, a = g = 9.8 m/s2
Therefore, 
               F = 0.05 x 10 = 0.5 N.
The magnitude of net force = 0.5 N, which is acting vertically in the downward direction.
If the pebble is thrown at an angle of 45° with the horizontal direction, it will have both the horizontal and vertical components of velocity. At the highest point, only the vertical component of velocity becomes zero. However, the pebble will have the horizontal component of velocity throughout its motion. This component of velocity produces no effect on the net force acting on the pebble. 

Question
CBSEENPH11020061
Wired Faculty App

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.

Solution

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.

Question
CBSEENPH11020062
Wired Faculty App

One end of a string of length l is connected to a particle of mass m and the other to a small peg on a smooth horizontal table. If the particle moves in a circle with speed v the net force on the particle (directed towards the centre) is:

(i) T, (ii) T - mv2 / l, (iii) T + mv2 / l, (iv) 0,

T is the tension in the string. 

[Choose the correct alternative]. 

Solution

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, 
Error converting from MathML to accessible text.
where, 
F is the net force acting on the particle. 

Question
CBSEENPH11020063
Wired Faculty App

A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with a speed of 15 ms–1. How long does the body take to stop?

Solution

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 = fraction numerator negative 50 over denominator space 20 end fraction space equals space minus 2.5 space m divided by s squared
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 = negative straight u over straight a space equals space fraction numerator negative 15 over denominator negative 2.5 end fraction space equals space 6 space s

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