Question

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

Answer 1

Answer:

t = 2.5 sec

#Answered@C29

Answer 2

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 :

$J=m(v-u)$

$Ft=m(v-u)$

$t=\dfrac{m(v-u)}{F}$

$t=\dfrac{m(-u)}{F}$

$t=\dfrac{5\times 5}{10}$

t = 2.5 seconds

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

Learn more,

Impulse

https://brainly.in/question/11580371