A 2 kg block is released from the top of a curved incline in theshape of a quart

Question

A 2 kg block is released from the top of a curved incline in theshape of a quarter of a circle radius R. The block then slides ontoa horizontal plane where it finally comes to rest 8 meters from thebeginning of the plane. The curved incline is frictionless, butthere is an 8-newton force of friction on the block while it slideshorizontally. assume g= 10m/s2
a. Determine the magnitude of the acceleration of the block whileit slides along the horizontal plane.
b. What time elapses while the block is sliding horizontally?
c. Calculate the radius of the incline in meters.

Explanation / Answer

Applying the law of conservation of energy we have, PE=KE mgh=1/2*mv2 v=2gh v=2gR The velocity at the bottom is v Th acceleration produced on the horizontal plane is, fr=ma 8=2*a a=4m/s2 The initial velocity is as v2-u2=2as 0-2gR=2*-4*8 R=3.265m v=26.1224m/s This is the radius of circle of it. We have is that as, v=u+at 0=26.1224-4*t t=6.5306s This is the time elapsed by it. v=2gh v=2gR The velocity at the bottom is v Th acceleration produced on the horizontal plane is, fr=ma 8=2*a a=4m/s2 The initial velocity is as v2-u2=2as 0-2gR=2*-4*8 R=3.265m v=26.1224m/s This is the radius of circle of it. We have is that as, v=u+at 0=26.1224-4*t t=6.5306s This is the time elapsed by it.

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