A block slides with constant velocity down a plane inclined at angle alpha to th

Question

A block slides with constant velocity down a plane inclined at angle alpha to the horizontal. If the block is projected up the same plane with initial velocity, v0, how far up the plane will it move before stopping? An engineer who weighs 160 lb stands on a platform that weighs 80 lb. She pulls a rope fastened to the platform and running over a pulley on the ceiling. What force must she exert to give the platform (and herself on the platform) an upward acceleration of 2.0 ft/s2? A block weighing 5 lb, connected to an unknown weight, W, is observed to slide down a 53degree inclined plane with an acceleration 4 ft/s2. When weight W is doubled, the block slides up the plane with acceleration 4 ft/s2. (see diagram to the right) Calculate weight W. What is the coefficient of sliding friction between the block and the plane? A block of mass 200 gm rests on top of mass 800 gm. The combination moves at constant velocity on a level surface with a hanging mass of 200 gm (see diagram). Find the accelaration of the system (diagram at right) when the 200gm block is removed from the 800 gm block and attached to the 200 gm hanging mass. What is the tension in the cord attached to the 800 gm block in the diagram at the right?

Explanation / Answer

here the data given is when the block is moving down wards ?=530 aceleration=4ft/sec2 here the formula for down ward acc=g(sin?-µcos?) 4=32(sin53-µcos53) 4/32=0.7986-µ 0.6018-----1 similarly here the formula for down ward acc=g(sin?-µcos?) 4=32(sin53-µcos53) 4/32=0.7986-µ 0.6018-----1 similarly when the block is moving up wards ?=530 aceleration=4ft/sec2 here the formula for up ward acc=g(sin?+µcos?) 4=32(sin53+µcos53) 4/32=0.7986+µ 0.6018-----2 solving eq 1 and 2 we have b) µ=0.2077 also a)weight of the block =5*4=20poundal when the block is moving up wards ?=530 aceleration=4ft/sec2 here the formula for up ward acc=g(sin?+µcos?) 4=32(sin53+µcos53) 4/32=0.7986+µ 0.6018-----2 solving eq 1 and 2 we have b) µ=0.2077 also a)weight of the block =5*4=20poundal also a)weight of the block =5*4=20poundal

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