Please solve this numerical

Asked by www1028 | 26th Aug, 2009, 09:53: PM

Expert Answer:

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

Answered by  | 27th Aug, 2009, 05:02: PM

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