Question

a stone of 1 kg is falling in front of a window of height 2m .if the velocity of stonr is 4m/sec then find the change in k.e. when it reaches to the bottom of window

Answer 1

Explanation:

Initial case:-

v=4 m/sec

m=1 kg

since,

K.E= 1/2 mv²

= 1/2× 1×(4)²

= 1/2× 16

= 8 J

Now final case i.e at the bottom of window:-

we have to find the final velocity of the partical to figure out final KE.

here,

u=4 m/sec

v=?

a=g=10 m/sec²

s=h=2 m

Now, Use equation of motion,

v²=u²+2as

=(4)² + 2×10×2

= 16 + 40

= 56 m/s

Hence , final

K.E = 1/2mv²

= 1/2 × 1× 56 [v²=56]

= 28 J

So the total change in K.E= 28J - 8J

= 20J.