Question

An object of mass 2kg is dropped from a height of 20 m above ground. Find the change in its kinetic
energy during the last Im of its fall towards ground. [Take g = 10 m/s)
(a) 15J
(b) 40J
(C) -35J
(d) 20J​

Answer 1

Velocity at last 1 metre= v2-u2=2as
Here u=0 a=10m/s2 S=19m
V2=0+2*10*19
V2=380

Initial Kinetic energy= 1/2mv2
(KE)1=1/2*2*380
(KE)1=380J

Velocity after last 1m
V2=0+2*10*20
V2=400

Final Kinetic Energy
(KE)2=1/2*2*400
(KE)2=400J

Therefore, change in Kinetic Energy=
KE final(KE2)-KE initial(KE1)
Change in KE=400-380
Change in KE=20J