Question

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

Answer 1

Part a)

since gravitational force is a conservative force so work done by conservative force in round trip is always ZERO

so in round trip work done by gravity must be ZERO

Part b)

Applied force while going upwards = force due to gravity along incline plane + force due to friction

$F = mgsin\theta + \mu mg cos\theta$

here we have

$\theta = sin^{-1}\frac{h}{L} = sin^{-1}\frac{5}{10} = 30 degree$

now we have

$F = 0.3*10*sin30 + 0.15*0.3*10*cos30$

$F = 1.89 N$

now work done to move it up

$W = F.d = 1.89 * 10 = 18.9 J$

so external work done to move it up is 18.9 J

Part c)

Work done by friction in round trip = force of friction * total distance

here we have

$F_f = \mu * mg cos30 = 0.15*0.3*10*cos30 = 0.39 N$

total distance of round trip = L + L = 10 + 10 = 20 m

now work done

$W = - 0.39 * 20 = - 7.79 J$

so work done by friction force will be - 7.79 J

Answer 2

Answer:

Here is your answer now go for it