A body of mass 1.5kg initially at rest is moved by horizontal force of 0.5Non a smoth frictionlss table. Calculate the work done by the force in 10sec .show that this is equal to the change in kinetic energy of body????
Answers
Answered by
1
we know that
work done =force /time
so W=0.5/10
= 0.05 joule
work done =force /time
so W=0.5/10
= 0.05 joule
Answered by
1
u=0
mass m=1.5kg
force acting on it F=0.5N
the acceleration of the body a=F/m=0.5/1.5=0.33m/s2
again, a=(v-u)/t or,velocity v=at=0.33*10m/s=3.3m/s
when t=10s
the distance covered by the body be s
v2=2as
s=v2/2a=16.5m
the work done W=F.s=0.5*16.5=8.25J
the change in kinetic energy of the body ∆ k=mv^2/2=1.5*(3.3)^2/2=8.25j
mass m=1.5kg
force acting on it F=0.5N
the acceleration of the body a=F/m=0.5/1.5=0.33m/s2
again, a=(v-u)/t or,velocity v=at=0.33*10m/s=3.3m/s
when t=10s
the distance covered by the body be s
v2=2as
s=v2/2a=16.5m
the work done W=F.s=0.5*16.5=8.25J
the change in kinetic energy of the body ∆ k=mv^2/2=1.5*(3.3)^2/2=8.25j
Similar questions