Question

a body of mass of 2 kg falls from rest. what will be it's kinetic energy during the fall at the end of 2 seconds?take g=9.8 metres per second square?

Answer 1

Given: m=2kg, u=0, t=2s, g=9.8m/s^2
w.k.t v=u+at
as u=0 and a= -g
=>v=0-gt
v=-9.8×2
v=-19.6m/s
kinetic energy (k)=1/2(mv^2)
k=1/2[2×(19.6)^2]
=384.16 kgm/s^2




hope it helps you....

Answer 2

Find velocity after 3 second by using equation of motion: v = u + at
v = 0 m/s + (9.8 m/s² × 2 s)
= 19.6 m/s

Kinetic energy is given as
K = 0.5 mv²
= 0.5 × 2 kg × (19.6 m/s)²
= 384.16 joules

∴ Kinetic energy of body after 2 s is 384.16 joules