7-27 A block of mass 2 kg is released from rest at point A on a track which is o

Question

7-27

A block of mass 2 kg is released from rest at point A on a track which is one quadrant of a circle of radius 1 m (Fig. 7-21). It slides down the track and reaches point B with a velocity of 4 ms^-1. From point B it slides on a level surface a distance of 3 m to point C, where it comes to rest. a) What was the coefficient of sliding friction on the horizontal surface? b) How much work was done against friction as the body slid down the circular arc from A to B? A small body of mass m slides without friction around the loop-the-loop apparatus shown in Fig. 7-22. It starts from rest at point A at a height 3R above the bottom of the loop. When it reaches point B at the end of a horizontal diameter of the loop, compute a) its radial acceleration, b) its tangential acceleration, and c) its resultant acceleration. Show these accelerations in a diagram, approximately to scale.

Explanation / Answer

7-27)Given,

m = 2 kg ; v(B) = 4 m/s ; d = 3 m

At point B, the block has kinetic energy which gets dissipated as thermal energy due to friction as it proceeds towards C and stops at C.

1/2 m v(B)^2 - F(fric) x d = 1/2 m v(C)^2

1/2 x 2 x 4^2 - F(fric) x 3 = 0

F(fric) = 5.33 N

We know that, F(frc) = uN

F(fric) = u x mg

u = F(fric)/mg = 5.33/2 x 9.8 = 0.272

Hence, u = 0.272

b)At A, the block has potential energy

PE(A) = m g h = 2 x 9.8 x 1 = 19.6 J

At B it has KE

KE(B) = 1/2 m v(B)^2 = 0.5 x 2 x 4^2 = 16 J

So the enrgy loss in transit due to friction is:

E = 19.6 - 16 = 3.6 J

Hence, E = 3.6 J

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