A 0.50 kg block is released from rest at the top of a 1.50 mlong smooth 36.9 deg

Question

A 0.50 kg block is released from rest at the top of a 1.50 mlong smooth 36.9 degrees inclined plane. A) find the wight of the block B.) find the normal (surface) force acting on the block. C.)Find the acceleration of the block D.)Find the time it takes for the block to reach the bottom ofthe incline. A 0.50 kg block is released from rest at the top of a 1.50 mlong smooth 36.9 degrees inclined plane. A) find the wight of the block B.) find the normal (surface) force acting on the block. C.)Find the acceleration of the block D.)Find the time it takes for the block to reach the bottom ofthe incline.

Explanation / Answer

mass m = 0.5 kg angle = 36.9 o length of the surface L = 1.5 m Height if the plane h = L sin (A). Weight of the block W = m g (B ). the normal (surface) force acting on the block = m g cos (C ). the acceleration of the block a = g sin (D ) . velocity of the block when it reaches theground   v = [ 2 gh ] from the relation v = u + at time taken   t = ( v - u ) / a where   u = initial velocity = 0 m / s substitue values weget answers

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