Question

A body moving at 2 m/s can be stopped over a distance x.If its kinetic energy is doubled,how long will it go before coming to rest,if the retarding force remains unchanged?
(a) x (b) 2x (c) 4x (d) 8x

Answer 1

S=v²/ 2a ......now is kE becomes 2 times velocity becomes √2 times thus stopping distance =2x

Answer 2

Hey dear,

◆ Answer-
x' = 2x

◆ Explanation-
# Given-
u = 2 m/s
v = 0 m/s
KE' = 2KE

# Solution-
As retarding force remains unchanged.
F = ma = constant
Therefore, a is constant.

Let u be initial velocity and u' be that after doubling KE.
KE' = 2KE
1/2 mu'2 = 2 × 1/2 mu^2
u'^2 = 2u^2

Before doubling KE,
v^2 = u^2 + 2ax
0 = u^2 + 2ax
x = -u^2 / 2a

After doubling KE,
v^2 = u'^2 + 2ax'
0 = u'2 + 2ax'
x' = u'^2 / 2a
x' = 2u'^2 / 2a
x' = 2x

A body will drag along 2x distance during stopping after doubling KE.

Hope this helps...