A 1kg block is released from from rest at point A as shown above, a vertical dis

Question

A 1kg block is released from from rest at point A as shown above, a vertical distance h above the ground. It slides down an inclined track, around a circular loop of radius .5m, then up another incline that forms an angle of 30 degrees with the horizontal. The block slides down off the track with a speed of 4 m/s at point C, which is a heght of .5m above the ground. Assume the entire track to be frictionless and air resistance to be negligible.

4. another track that has the same configurations, but is not frictionless, is used. with this track it is found that if the block is to reach point C with a speed of 4 m/s, the height h must be 2m. Determine the work done by the frictional force.

Explanation / Answer

I guess the first question is to find h (it does not say so explicitly anywhere above). Since the track is frictionless, the hoop is irrelevant, whatever energy the block gains when going up is lost coming down. The inclination is also irrelevant. Use conservation of energy: At top of ramp (of height = h) E=mgh At top of the other ramp, at height = 0.5: E=mg*0.5 + 0.5*m*(4^2). (energy = potential + kinetic =mg*height + 0.5 m velocity^2) Equating E from the two equations, gh = 0.5g + 8; h=0.5 + (8/g) = 1.3 metres assuming g=10 m/s^2 (b) Clearly whatever energy the block would have gained by falling from height h=2 m to 1.3 m if the track were frictioless, is lost due to friction, there being no other difference between the two tracks or between the final velocities on both tracks. This difference = mg*(2-1.3) = 1*10*0.7 = 7 Joules. This is the work done by frictional forces, again assuming g=10 m/s^2

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