A 10.0 kg block is released from point A in the figure below. The track is frict
ID: 1410202 • Letter: A
Question
A 10.0 kg block is released from point A in the figure below. The track is frictionless except for the portion between points B and C, which has a length of 6.00 m. The block travels down the track, hits a spring of force constant 2250 N/m, and compresses the spring to 0.300 m from its equilibrium position before coming to rest momentarily. Determine the coefficient of kinetic friction between the block and the rough surface between B and C. A railroad car of mass 2.55 times 10^4 kg, is moving with a speed of 4.20 m/s. It collides and couples with three other coupled railroad cars, each of the same mass as the single car and moving in the same direction with an initial speed of 2.10 m/s. What is the speed of the four cars after the collision? How much energy is transformed into heat, light and sound in the collision?Explanation / Answer
1. initial energy Ei = mgh
final energy Ef = ½kx²
work done by friction W = mgdcos(180) = -mgd
W = Ef - Ei
-mgd = ½kx² - mgh
so
= h/d - kx²/(2mgd)
= 3.00/6.00 - 2250*0.300²/(2*10.0*9.81*6.00)
= 0.327
2. If you use conservation of momentum, you can find that :
(m1v1 + m2v2)i = (m1v1 + m2v2)f
(2.55*10^4)(4.20) + (2.55*10^4*3)(2.10) = (2.55*10^4*4)( vf )
By the way, in the latter half of the equation, I used ( 2.43104 * 4 ) because they couple together, therefore having 4 times the weight for 4 cars.
Therefore if you solve for Vf, you get :
Vf = 2.625 m/s
Question b)
If you use Kinetic Energy, because it isn't conserved, you get:
1/2 mv^2 + 1/2 mv^2 = 1/2 mv^2 + internal energy
(1/2) ( 2.55*10^4) ( 4.20)^2 + (1/2) ( 2.55*10^4 * 3 ) (2.10)^2 = (1/2) ( 2.55*10^4 * 4 ) (2.625) ^2 + internal energy
Solving for internal energy:
Internal Energy = 39.35 - 35.14 = 4.21J
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