The 0.2 kg box below slides down a curved ramp, jumps a small gap and lands on a
ID: 1492108 • Letter: T
Question
The 0.2 kg box below slides down a curved ramp, jumps a small gap and lands on a flat platform. At the point on the ramp shown it is 1.5 m above the floor and its speed is 2.0 m/s. At the point shown on the platform the box is 0.4 m above the floor and sliding at 4.2 m/s. If we consider a system consisting of the box and Earth's gravity field so that E = K + U_g, has the energy of this system been conserved during the described process? Explain how you know. If we consider the exact same process but broaden our system definition so that E = K + U_g + E_other and E_other includes any "other" form of energy that might have been produced through the process (most of it is thermal), what objects are included in this system? Discuss, don't just state a list. Determine deltaE_other for the process as described. How much kinetic energy would the box have on the platform if deltaE_other = 0 ?Explanation / Answer
a) Ei = Ki+Ui = 1/2mvi^ + mghi = 1/2*0.2*2.0^2 + 0.2*9.8*1.5 = 3.34 J
Ef = Kf+Uf = 1/2mvf^ + mghf = 1/2*0.2*4.2^2 + 0.2*9.8*0.4 = 2.55 J
By law of conservation of energy, to conserve energy Ef=Ei
Since Ef > Ei energy is not conserved.
c)
Friction force might be associated between the surface of the block and the ramp. Due to friction some energy is lost.
Ei = Ki+Ui
Ef = Kf+Uf + Eother
By law of conservation of energy
Ef= Ei
Kf+Uf + Eother = Ki+Ui
Eother = (Kf+Uf) – (Ki+Ui) = 2.55 – 3.34 = - 0.79 J
d)
If Eother = 0
(Kf+Uf) = (Ki+Ui)
1/2mvf^2 + mghf = 1/2mvi^ + mghi
1/2*0.2*vf^2 + 0.2*9.8*0.4 = 1/2*0.2*2.0^2 + 0.2*9.8*1.5
Gives, vf= 5.06 m/s
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