A 500 g block is released from the top of a 45° incline. The block is initially

Question

A 500 g block is released from the top of a 45° incline. The block is initially at a distance of D=80 cm from the free end of a long massless spring that lies on the incline’s surface with its other end held fixed at the bottom edge of the incline. The unstretched length of the spring is L= 17 cm. The block slides down the incline and comes to a momentary stop at the maximum compression of the spring when the spring’s length becomes 13 cm. The spring then rebounds and the box begins to move back up the plane. The static and kinetic coefficients of friction between the block and the incline are equal µk = µs = 0.6. Find the total energy loss due to friction over the entire way from the initial location of release all the way to the highest position the block attains after the rebound.

Explanation / Answer

given,

mass = 500 g

angle = 45 degree

D = 80 cm

L = 17 cm

x = 17 - 13

x = 4 cm

µk = µs = 0.6

initial energy = 0.5 * 9.8 * 0.8 * sin(45)

initial energy = 2.771 J

energy lost in downward motion = 0.6 * 0.5 * 9.8 * 0.8 * cos(45)

energy lost in downward motion = 1.663 J

energy left = 2.771 - 1.663

energy left = 1.108 J

1.108 = 0.6 * 0.5 * 9.8 * x * cos(45) +  0.5 * 9.8 * x * sin(45)

x = 0.1998

energy lost = 1.663 + 0.5 * 9.8 * 0.1998 * sin(45)

energy lost = 2.35 J

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