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 this information to answer the questions below:

hA = 10.0 m, the height at "A"

hB = 2.50 m, the height at "B"

vA = 2.00 m/s, the initial speed of the cart at "A"

vC = 11 m/s, the speed of the cart at "C"

m = 8.00 kg, the mass of the cart

(A) What speed should the Joker expect to measure when the cart is at "B?"

vB = _____ m/s

(B) What is the height at "C?"

hC = _____ m

(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 9.23. How much energy was lost by the cart due to the friction?

ELost = _____ J

Please show work.

Explanation / Answer

A)Total Energy at A is Total Energy at B

(KE+PE)A=(KE+PE)B

0.5 MVA^2+MgHA = 0.5 MVB^2+MgHB

0.5 VA^2+gHA = 0.5 VB^2+gHB

0.5 VA^2+gHA -gHB = 0.5 VB^2

on substituting given values, 0.5 VB^2 = 75.5 , So, VB =12.288 m/s2

B)

We need to calculate the height at C, We use the same principle

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

0.5 MVA^2+MgHA = 0.5 MVC^2+MgHC

0.5 VA^2+gHA = 0.5 VC^2+gHC

gHC =0.5 VA^2+gHA -0.5 VC^2 = 39.5

HC = 39.5/g = 4.03 m

C) Total energy without friction is

0.5 MVB^2+MgHB where VB is 12.288 m/s

Total Energy = 799.97 J

WIth friction resulting in new velocity as VB = 9.23 m/s, total energy is

Energy with friction is Ef = 536.77 J

Energy lost due to friction is 263.19 J

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