Answer to Question #97849 in Mechanics | Relativity for Stefanie

Question

As shown in the figure, a 0.540 kg container is pushed against a horizontal spring of negligible mass until the spring is compressed a distance x. The force constant of the spring is 450 N/m. When it is released, the container travels along a frictionless, horizontal surface to point A, the bottom of a vertical circular track of radius
R = 1.00 m,
and continues to move up the track. The speed of the container at the bottom of the track is vA = 12.3 m/s, and the container experiences an average frictional force of 7.00 N while sliding up the track.
a. What is x?
b. If the container were to reach the top of the track, what would be its speed (in m/s) at that point?
c. Does the container actually reach the top of the track, or does it fall off before reaching the top?
reaches the top of the track or falls off before reaching the top or not enough information to tell

Answer 1

a. The compression can be found from the law of conservation of energy:

$$\frac{1}{2}mv^2=\frac{1}{2}kx^2,\\ \space\\ x=v\sqrt{m/k}=0.43\text{ m}.$$

b. The kinetic energy will be spent to increase the potential energy up to the height 2R and to overcome the friction along half of the circle:

$$\frac{1}{2}m(v_{\text{bot}}^2-v^2_{\text{top}})=mg\cdot2R+fd,\\ \space\\ d=2\pi R,\\ \space\\ v_{\text{top}}=\sqrt{v_{\text{bot}}^2-4gR-\frac{4\pi fR}{m}}<0,$$

that is why the speed at the top cannot be found.

c. As we see from the previous part, the container will fall of before reaching the top.