As shown in the figure, a 1.45-kg block is held in place against the spring by a

Question

As shown in the figure, a 1.45-kg block is held in place against the spring by a 21-N horizontal external force. The external force is removed, and the block is projected with a velocity v 1=1.2 m/s as it separates from the spring. The block descends a ramp and has a velocity v 2=2.1 m/s at the bottom. The track is frictionless between points A and B. The block enters a rough section at B, extending to E. The coefficient of kinetic friction between the block and the rough surface is 0.29. The velocity of the block is v 3=1.4 m/s at C. The block moves on to D, where it stops. How much work is done by friction between points B and C?

Explanation / Answer

initial velocity at B = Vi = 2.1 m/s

final velocity at C = Vf = 1.4 m/s

m = mass of block = 1.45 kg

the change in kinetic energy is given as ::

change in kE = KEf - KEi = (0.5) mVf2 - (0.5) mVi2

change in kE = (0.5) (1.45) (1.4)2 - (0.5) (1.45) (2.1)2 = - 1.78 J

according to work-kinetic energy theorem ::

work done = change in KE = -1.78 J

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