Answer to Question #160517 in Physics for brayden

Question

A small block, with a mass of 4 kg, starts from rest at the top of the apparatus shown above. It then slides without friction down the incline, across the first level section, around the loop and then onto the final level section on the right. It then collides with a spring and compresses it, which momentarily brings the block to a stop. The maximum height of the incline is 5 m, the radius of the loop is 1.0 m and the spring constant is 120 N/m. Find the initial potential energy of the block at point A. Find the velocity the block at the bottom of the incline at point B.Find the velocity of the block at the top of the loop at point C.How much will the block compress the spring before coming to a stop at point D?

Answer 1

a) From the definition of the potential energy, we get:

$$PE_i=mgh=4\ kg\cdot9.8\ kg\cdot5\ m=196\ J.$$

b) We can find the velocity of the block at the bottom of the incline at point B from the law of conservation of energy:

$$PE_i+KE_i=PE_f+KE_f,$$ $$PE_i+0=0+KE_f,$$ $$mgh=\dfrac{1}{2}mv^2,$$ $$v=\sqrt{2gh}=\sqrt{2\cdot9.8\ \dfrac{m}{s^2}\cdot5\ m}=9.89\ \dfrac{m}{s}.$$

c) We can find the velocity of the block at the top of the loop at point C from the law of conservation of energy:

$$PE_i+KE_i=PE_f+KE_f,$$ $$PE_i+0=PE_f+KE_f,$$ $$mgh_{inc}=mgh_{loop}+\dfrac{1}{2}mv^2,$$ $$v=\sqrt{2g(h_{inc}-h_{loop})}=\sqrt{2\cdot9.8\ \dfrac{m}{s^2}\cdot(5\ m-2\cdot1\ m)}=7.67\dfrac{m}{s}.$$

d) We can find how much will the block compress the spring before coming to a stop at point D from the law of conservation of energy:

$$PE_i+KE_i=PE_f+KE_f,$$ $$PE_i+0=PE_f+0,$$ $$PE_i=\dfrac{1}{2}kx^2,$$ $$x=\sqrt{\dfrac{2PE_i}{k}}=\sqrt{\dfrac{2\cdot196\ J}{120\ \dfrac{N}{m}}}=1.8\ m.$$