Question

A body of mass 62 kg comes to rest after moving a distance of 20 m in a time interval of

5 seconds. Find the work done.​

Answer 1

From above figure

A→B

u = 4m/s ,v=0 , t=2sec, a=?

v=u+at

0=4+a(2)

a=−2m/s²

so,

force = ma

=2×2=4N

The Answer is 4 N

Answer 2

Given, mass = 62Kg

v = 0

s = 20m

t = 5 s

To find a

S = ut + 1/2 at^2

20 = 5u + 1/2 a 25

20 - 12.5a = 5u

u = 4 - 5/2 a

V = u +at

0 = 4 - 2.5 a + 5a

a = (4)/(5/2)

a = 8/5 m/s^2

F = ma

F = 62 × 8/5

F = 12.4 × 8

F = 99.2 N

therefore Work Done = F•S

Work = 99.2 × 20

Work = 1984 J