CBSE Class 11-science Answered
Given: m = 0.3 kg, h = 5 m, length (l) = 10 m;
(a) work done by conservative force in a round trip is always zero.
(b) if theta is the angle at which the plane is inclined, then
sin (theta) = 5 / 10 = 1/2
cos (theta) = sqrt (3) / 2
Applied force = frictional force + parallel component of weight
= (mu) R + mg sin (theta) ; where R = mg cos(theta)
Work done = applied force x distance
= [(mu) mg cos(theta) + mg sin(theta) ] x distance
= mg [(mu) cos (theta) + sin (thata)] x distance
= 0.3 x 9.8 [0.15 x sqrt (3) / 2 + 1/2] x 10
= 18.5 J
(c) Work done by the frictional force over the round trip = - (mu) mg cos(theta) x S - (mu) mg cos(theta) x S
After substituting the values of parameters, we have
Work done by the frictional force over the round trip = - 7.63 J
(d) Kinetic energy of the body at the end of the trip = total work done by the applied force and the frictional force
= 18.5 - 7.6 = 10.9 J