Question

A mass of 40 kg was moving with a velocity 200 m/s. A force of 20000 N is applied on the mass and its velocity is reduced to 100 m/s after some time. What is the distance travelled by the mass during this period ?

Answer 1

F=Ma

20,000 = 40 × a

a = 20,000 ÷ 40

a = 500 ms^-2

we applied acceleration in the opposite direction therefore it should be of negative and becomes retardation

a = -500m/s^2

now, we use the formula

v^2 - u^2 = 2as

we have ; u (initial velocity) = 200 m/s

v (final velocity) = 100 m/s

a (acceleration) = -500 m/s^2

now by putting values in the above formula, we get

(100)^2 - (200)^2 = 2 (-500) s

10,000 - 40,000 = -1000 × s

-30,000 = -1000 × s

s= -30,000 ÷ -1000

s= 30m

HOPE IT HELPS