Question

The 5kg cabinet is movies initially at a velocity 5 m/s. The impulse force of the wall on the cabinet to stop is 10 N. Neglect friction between cabinet and wall. Calculate the time of cabinet to stop after impulse?

Answer 1

Answer:

The time of cabinet to

stop after impulse is

2.5 seconds

Explanation:

Given that,

Mass of the cabinet, m = 5 kg

Initial speed of the cabinet, u = 5 m/s

Finally, it stops, v = 0

Force of the wall on the cabinet, F = 10 N

We know that the impulse is equal to the change in momentum such that :

$j = m(v - u)$

$ft = m(v - u)$

$t = m(v - u) \div f$

$t = m ( - u) \div f$

$t = 5 \times 5 \div 10$

$t = 2.5sec$

t = 2.5 seconds

So, the time of cabinet to stop after impulse is 2.5 seconds. Hence, this is the required solution.