Derive the expression of kinetic energy
b) A body of mass 50 kg varies its velocity from 3m/s to 5m/s in 5 seconds. What
is the work done by the body?
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b) According to the work energy theorem,
For work done W and kinetic energy ke
W=∆ke
W=ke(f)-ke(i)
=Mv(f)^2÷2 - Mv(i)^2÷2
={M(v(f)^2 - v(i)^2)}÷2
={10(3^2 -5^2)}÷2
={10(9–25)}÷2
=(10×-16)÷2
=-80 joules
As the work is negative in value means that the work is done by the body and it is of 80 joules.
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