1. The potential energy due to gravitation is V(h)=mgh. Consider the problem of

Question

1. The potential energy due to gravitation is V(h)=mgh. Consider the problem of a ball of mass rolling on a one-dimensional hill that is described by the equation , h=(x-5)^2 where stands for the position.

(a) Determine an expression for the potential energy of the ball as a function for the position .

(b) Determine the position where the potential energy is minimized.

(c) Determine the change in potential energy that accompanies the motion of the ball starting at rest (velocity v=0) at x=0 to the minimum energy position you determined in part Provide an interpretation for the sign of the energy change.

(d) If the motion in part (c) occurs without friction, how fast is the ball moving at the minimum energy position?

(e) What happens after the ball reaches the minimum energy position? Explain.

(f) Using the equation for the potential energy due to gravitation from problem 1, derive an expression for the force.

(g) Determine the work required to lift a 1 kg mass from the basement of Chemistry up to the fifth floor; the floor to floor distance is approximately 12 feet = 3.66 m.

Explanation / Answer

a) V=mg(x-5)^2

b)dV/dx = mg*2(x-5) =0

x=5

c)At x=0, V1=25mg

V2=0

del V= -25mg

d) velocity =(x-5) (2g)^1/2

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