A block of mass 0.500 kg is pushed against an ideal spring until the spring is c

Question

A block of mass 0.500 kg is pushed against an ideal spring until the spring is compressed a distance x. The force constant of the spring is 350 N/m. When it is released, the block travels along a frictionless, horizontal surface to point B at the bottom of a circular track of radius 1.30 m. The speed at B is vB= 11.8 m/s.

(a) What is x?

(b) The block encounters an average frictional force of 5.30 N as it slides up the track. Use an energy

analysis to determine the speed at the top.

(c) Show that this speed is less than the minimum speed required for the block to stay on the track at

the top. (Incidentally, this shows that the block cannot in fact make it to the top under the given

conditions.)

(d) What is the minimum value of x for the block to be able to reach the top of the loop?

Explanation / Answer

(a) Applying energy conservation,

350 x^2 / 2 =0.500 * 11.8^2 / 2

x = 0.446 m

(b) Work done by friction = - (5.30) (pi x 1.30) = - 21.65 J

kinetic Energy at top = (0.500 * 11.8^2 /2 ) - 21.65 - (0.5x 9.81 x 2 x 1.30) = 0.5 v^2 /2

v = 1.27 m/s

(c) to remain in the loop,

N + m g = m v^2 / r

v = sqrt(r g) = sqrt(1.30 x 9.81) = 3.57 m/s this is rerquired speed.

so speed is less than required speed.

(d) Applying energy conservation,

(0.5 x 3.57^2 /2 ) + (0.5 x 9.81 x 2 x 1.30) + (21.65) = 350 x^2 /2

x = 0.463 m

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