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A body of mass 2 kg is thrown upward with initial velocity of 20 m/s. After 2 seconds, its kinetic energy will be : (g = 10 m/s2)
Answer & Solution
Correct Answer:
Option
A
We will use the following equation of motion to get the final velocity after 2 seconds.
V = u - gt
U = 20 m/s
g = 10
t = 2
V = 20 - 2 × 10
V = 20 - 20
V = 0
K.E = 1/2mv²
m = 2
v = 0
K.E = 0.5 × 0 × 2
= 0
V = u - gt
U = 20 m/s
g = 10
t = 2
V = 20 - 2 × 10
V = 20 - 20
V = 0
K.E = 1/2mv²
m = 2
v = 0
K.E = 0.5 × 0 × 2
= 0
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LoginMass of body ,m=2 kg
Initial velocity u=20m/s
Height reached at t=2secs
h=ut−1/2 gt^2
⟹h=20×2− 1/2( 10) (2)^2
===>H =40-20
===>h=20 m
Potential energy , P=mgh=2×10×20=400 J