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 the plane is 0.15. What is the friction between the body and the plane is 0.15. what is the a.) work done by the gravitational force over the round trip. b.) work done by the applied force over the round upward journey , c.) work done by the frictional force over the round trip ,d.) kinetic energy of the body at the end of the trip ?
Answers
Answer:
Given that
m = 0.3 kg
h = 10 m
P = 5 m
µ = 0.15
sin θ= CB/CA = 0.5
sin θ= CB/CA = 0.5 => θ= 30°
(i) W = FS = – mg sin q × h = – 14·7 J is the W.D. by gravitational force in moving the body up the inclined plane.
W′ = FS = + mg sin θ × h = 14·7 J is the
W.D. by gravitational force in moving the body down the inclined plane.
∴ Total W.D. round the trip,
W1 = W + W′ = 0. (Work done by gravitation force over the round trip = 0 As displacement is zero.)
(ii) Force needed to move the body up the inclined plane,
F = mg sin θ + fk
= mg sin θ + µk R
= mgsin θ + µk mg cos θ
∴ W.D. by force over the upward journey is
W2 = F × l
= mg (sin θ + µk cos θ) l
= 18·5 J
(iii) Work completed by frictional force over the round trip.
W (friction) = 2 f h
= 2 (µ mg cosθ. h)
= 2× 0.15×0.3×10×√75/10× 10
=> 7.79j
(iv) K.E. of the body at the end of round trip = W.D. by net force in moving the body down the inclined plane
= (mg sin θ – µk mg cos θ) l
= 10( 5/10-0.15×√75/10)
=> 10( 0.5 -0.13)
=> 10(0.37)
= 3.7 m/s^2
so speed at the end of trip
v^2 = u^2 + 2as
v^2 =0+ 2( 3.7) ×(10)
v^2 = 74 => 1v = 8.6 m/s
k.E = 1/2 mv^2
=> 0.5×0.3 × 74
=> 11.1J
⇒ K.E. of body = net W.D. on the body.
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@GauravSaxena01
