A man pulls a block of mass m = 23 kg up an incline at a slow constant velocity

Question

A man pulls a block of mass m = 23 kg up an incline at a slow constant velocity for a distance of d = 5.5 m. The incline makes an angle q = 29 with the horizontal. The coefficient of kinetic friction between the block and the inclined plane is k = 0.1.

At the top of the incline, the string breaks and the block, assumed to be at rest when the string breaks, slides down a distance d = 5.5 m before it reaches a frictionless horizontal surface. A spring is mounted horizontally on the frictionless surface with one end attached to a wall. The block hits the spring, compresses it a distance L = 0.5 m, then rebounds back from the spring, retraces its path along the horizontal surface, and climbs up the incline.

d) How far up the incline d1 does the block rebound?

Explanation / Answer

m=23 kg==>d=5.5 m==>=29o==>k=0.1==>L=0.5 m

change in pe=23*9.8*5.5*sin()601.02 Joules

Work=23*9.8*0.1*cos()*5.5108.43 Joules

difference in energy=601.02-108.43492.59 Joules

1/2*k*x2=492.59k=492.59/(5.52*0.5)32.568 N/m

m*cos()*g*k+m*sin()*g129 N

distance up slope=492.59/1293.82 m up incline

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