A block starts at rest and slides down a frictionless ramp. At the based of the

Question

A block starts at rest and slides down a frictionless ramp. At the based of the ramp is a circular loop-the-loop of radius 0.7 m. (Figure 1) Part A What is the minimum speed that the block must have at the top of the loop-the-loop so that it makes it around safely? (Speed = 2.62 m/s ) Part B What is the minimum height the block must start above the bottom of the loop-the-loop so that it makes it around the loop-the-loop safely? Height = m Have part A correct but can not figure out B i know B is not 1.4 or 4.8

Explanation / Answer

Since the first incline is frictionless,

there is no energy loss in coming down. Hence KE at the foot of the incline = PE at the start. This gives the velocity at the foot of the first incline as

1/2 m*v^2 = m*g*h.

= 2.62 as given

The frictional force on the horizontal surface = Uk*m*g

The work done in overcoming the fricional force =Uk*m*g*L

Thus the available energy at the foot of the second incline = m*g*h - Uk*m*g*L = m*g*(h - Uk*L)

If the height to which the block rises is h1,

m*g*h1 = m*g*(h - Uk*L). Hence

h1 = h - Uk*L

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