The Joker (another one of Batman\'s enemies) has a new toy: a spring-launched ca

Question

The Joker (another one of Batman's enemies) has a new toy: a spring-launched cart that delivers a high-explosive bomb underneath cars on street. The wheels use a unique air-bearing design in order to minimize friction. To test the wheels, Joker sets up the following experimental set up. The cart will begin at "A," with a certain velocity. Joker can then measure the velocities at "B" and "C. Use the information to answer the question below: h_A = 12.0 m, the height at "A" h_B = 3.00 m, the height at "B" V_A = 2.00 m/s, the initial speed of the cart at "A" V_C = 12.29 m/s, the speed of the cart at "C" m = 8.00 kg, the mass of the cart What speed should the Joker expect to measure when the cart is at "B?" What is the height at "C?" In one of Joker's trial runs, a wheel axle broke, creating a significant amount of friction. In that particular trial, the speed of the cart at "B" was 10.05. How much energy was lost by the cart due to the friction?

Explanation / Answer

A)

using conservation of energy between point A and B

PEA + KEA = PEB + KEB

mghA + (0.5) m VA2 = mghB + (0.5) m VB2

inserting the values

(9.8) (12) + (0.5) (2)2 = (9.8) (3) + (0.5) VB2

VB = 13.4 m/s

B)

using conservation of energy between point C and B

PEC + KEC = PEB + KEB

mghC + (0.5) m VC2 = mghB + (0.5) m VB2

inserting the values

(9.8) hC + (0.5) (12.29)2 = (9.8) (3) + (0.5) (13.43)2

hC = 4.49 m

C)

Energy lost to friction = Total energy at A - total energy at B

Energy lost to friction = PEA + KEA - PEB - KEB = mghA + (0.5) m VA2 - mghB - (0.5) m VB2

Energy lost to friction = (8)((9.8) (12) + (0.5) (2)2 - (9.8) (3) - (0.5) (10.05)2 )

Energy lost to friction = 318 J

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