A body of mass 500 gm is moving with a speed of 10 ms". A force acts on it
makes it move with a speed of 20 ms, g lo
Find (i) Change in K.E. of the body
ii) Work done by the force on the body.
Answers
Answered by
2
Answer:
Explanation:
Delta k =1/2m(20^2-10^2)
=1/4(300)
75
W =delta k=75J
Answered by
2
Answer:
HEY THERE!!
Given, m = 500 g = 0.5 kg, u = 10 ms^-1, v = 20 ms^-1
1) Increase in kinetic energy = 1/2 m (v^2-u^2)
=1/2×0.5×[(20)^2-(10)^2]
= 75 J
2) Work done by the force = increase in kinetic energy = 75 J
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