In the figure, a small block of mass m = 0.024 kg can slide along the frictionle

Question

In the figure, a small block of mass m = 0.024 kg can slide along the frictionless loop-the-loop, with loop radius R = 12 cm. The block is released from rest at point P, at height h = 3R above the bottom of the loop. How much work does the gravitational force do on the block as the block travels from point P to (a) point Q and (b) the top of the loop? If the gravitational potential energy of the block-Earth system is taken to be zero at the bottom of the loop, what is that potential energy when the block is (c) at point P, (d) at point Q, and (e) at the top of the loop?

Explanation / Answer

(a)

W = -mg(yf-yi)

=-mg*(R-3R)

=2*m*g*R

=2*0.024*9.8*12*10^-2

= 0.05645 J

(b)

W = -mg(yf-yi)

=-mg*(2R-3R)

=m*g*R

=0.024*9.8*12*10^-2

= 0.028224 J

(c)

U= 3*m*g*R

=3*0.024*9.8*12*10^-2

= 0.084672 J

(d)

U= m*g*R =0.024*9.8*12*10^-2

=0.028224 J

(e)

U= 2*m*g*R=2*0.024*9.8*12*10^-2

=0.056448 J

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