A small steel ball of mass \"m\" is released from rest at point A at height h on

Question

A small steel ball of mass "m" is released from rest at point A at height h on a smooth track. The ball is at height h_A and height h_c at C. Answer the following questions, using only fundamental constants and the given variables. What is the gravitational potential energy of the ball at point A? What is the kinetic energy of the ball at point D? What is the speed of the ball at point C? Is the speed of the ball at point B greater or less than the speed at point C? Explain. In a second trial, the ball is given an initial speed v_0 at point A as it is released. What is the new speed of the ball at point C?

Explanation / Answer

a) PE = m*g*hA

b) Using law of conservation of energy

KEi + PEi = KEf + PEf

KE(A) + PE(A) = KE(D) + PE(D)

1/2mvA2 + m*g*hA = KE(D) + m*g*hD

KE(D) = 1/2mvA2 + m*g*hA - m*g*hD

Since hA = hD and vA =0m/s

KE(D) = 0 J

c) Using law of conservation of energy

KEi + PEi = KEf + PEf

KE(A) + PE(A) = KE(C) + PE(C)

1/2mvA2 + m*g*hA = KE(C) + m*g*hC

KE(C) = 1/2mvA2 + m*g*hA - m*g*hC

KE(C) = 1/2mvA2 + m*g*(hA - hC)

Since vA =0m/s

KE(C) = m*g*hAC -------------------(1)

d) From equn(1) it is clear KE or the speed at a point is depend on change in height h. Hence since hAC < hAB , speed at B is greater than speed at C.

e) Using law of conservation of energy

KEi + PEi = KEf + PEf

KE(A) + PE(A) = KE(C) + PE(C)

1/2mvA2 + m*g*hA = 1/2mvC2 + m*g*hC

1/2mvC2 = 1/2mvA2 + m*g*hA - m*g*hC

1/2mvC2 = 1/2mvA2 + m*g*(hA - hC)

1/2mvC2 = 1/2mv02 + m*g*hAC

1/2vC2 = 1/2v02 + g*hAC

vC2 = v02 + 2g*hAC

vC = sqrt(v0 + 2g*hAC)

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