Advertisements
Advertisements
Question
A force acts for 10 s on a stationary body of mass 100 kg, after which the force ceases to act. The body moves through a distance of 100 m in the next 5 s. Calculate: The velocity acquired by the body.
Advertisements
Solution
Mass, m = 100 kg
Distance moved, s = 100 m
Initial velocity, u = 0
Because the body moves through a distance of 100 m in 5 s,
Velocity of the body = `"Distance moved"/"time taken"`
Velocity = `100/5`
= 20 m/s
APPEARS IN
RELATED QUESTIONS
Consider the Atwood machine of the previous problem. The larger mass is stopped for a moment, 2.0 s after the system is set into motion. Find the time that elapses before the string is tight again.
A constant force F = m2g/2 is applied on the block of mass m1 as shown in the following figure. The string and the pulley are light and the surface of the table is smooth. Find the acceleration of m1.
In the previous problem, suppose m2 = 2.0 kg and m3 = 3.0 kg. What should be the mass m, so that it remains at rest?
In the following figure shows a man of mass 60 kg standing on a light weighing machine kept in a box of mass 30 kg. The box is hanging from a pulley fixed to the ceiling by a light rope, the other end of which is held by the man himself. If the man manages to keep the box at rest, what is the weight recorded on the machine? What force should he exert on the rope to record his correct weight on the machine?
The unit of linear momentum is :
A force of 10 N acts on a body of mass 2 kg for 3 s, initially at rest. Calculate : The velocity acquired by the body
A motorcycle of mass 100 kg is running at 10 ms−1. If its engine develops an extra linear momentum of 2000 Ns, calculate the new velocity of a motorcycle.
Multiple Choice Question. Select the correct option.
Which of the following are vector quantities?
A stone is dropped from a cliff 98 m high.
How long will it take to fall to the foot of the cliff?
A cricket ball of mass 150 g has an initial velocity `u = (3hati + 4hatj)` m s−1 and a final velocity `v = - (3hati + 4hatj)` m s−1 after being hit. The change in momentum (final momentum-initial momentum) is (in kg m s1)
