A body of mass 0.3 kg is taken up an inclined plane of length 10 m and height 5 m , and then allowed to slide down to the bottom again. the coefficient of friction between the body and he plane is 0.15 . what is the
- work done by the gravitational force over the round trip ,
- wok done by the applied force over the upward journey,
- work done by frictional force over the round trip,
- kinetic energy of the body at the end of the trip?
how is the answer to d related to first three answers?
please answer my question fast ...................... experts of meritnationplease explain it elaborately and in an easy way please answer it fast
Hello,
Given that
m = 0.3 kg
h = 10 m
P = 5 m
µ = 0.15
(a) Work done by gravitation force over the round trip = 0
As displacement is zero.
(b) Work done by the applied force over the upward journey
Wup = Force × displacement
= (mg sinθ + f) h (As the block moves up friction would be downward)
= (mg sinθ + µ mg cosθ) h [since f = µ N = µ mg cosθ]
= mg (sinθ + µ cosθ) × h
(c) Work done by frictional force over the round trip.
W (friction) = 2 f h
= 2 (µ mg cosθ. h)
(d) Acceleration of the block.
Apply Newton's 2nd law
(mg sinθ – f) = ma
⇒ mg sinθ – µ mg cosθ = ma
⇒ a = g sinθ – µg cosθ
So speed at the end of trip
