Problem #1
The roller-coaster car shown reaches a vertical height of only 25 m on the second hill before coming to a momentary stop. It traveled a total distance of 400 m. Determine the thermal energy produced and estimate the average friction force (assume it is roughly constant) on the car, whose mass is 1000 kg. |
| Fig.1 |
The roller-coaster car is released at point 1 so its initial speed (and hence, kinetic energy) is zero: Ki = 0. If we measure height upwards from the level part of the track, then the initial potential energy for the mass (all of it gravitational) is
Ui = mgy1 = (1000.0 kg)(9.80 m/s2)(40.0 m) = 3.92 × 105 J
Next, for the “final” position of the roller-coaster car (This is shown in Fig. 1.) We don’t need to think about what the mass was doing in between these two points; we don’t care about the speed of the mass during its slide.
At this final point, the mass is again at rest, so its kinetic energy is zero: Kf = 0. Being at height y2 = 25 m, then the final potential energy for the mass (all of it gravitational) is
Uf = mgy2 = (1000.0 kg)(9.80 m/s2)(25.0 m) = 2.45 × 105 J
The thermal energy (ETH) produced is equal to the loss in mechanical energy.
ΔK +ΔU = Wnon-konc = –ETH
Kf – Ki + Uf – Ui = Wnon-konc
0 – 0 + 2.45 x 105 J – 3.92 x 105 J = –ETH
ETH = 1.47 x 105 J
Work by friction is equal to the change in mechanical energy
WNC = –fkd = –ETH
fk = ETH/d = 1.47 x 105 J/400 m = 370 N
Problem #2
The coefficient of friction between the 3.0kg mass and surface in Fig. 2 is 0.40. The system starts from rest. What is the speed of the 5.0kg mass when it has fallen 1.5m?
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| Fig.2 |
Answer:
When the system starts to move, both masses accelerate; because the masses are connected by a string, they always have the same speed. The block (m) slides on the rough surface, and friction does work on it. Since its height does not change, its potential energy does not change, but its kinetic energy increases. The hanging mass (M) drops freely; its potential energy decreases but its kinetic energy increases.
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| Fig.3 |
The forces acting on m are shown in Fig. 3(a). The normal force N must be equal to mg, so the force of kinetic friction on m has magnitude μkN = μkmg. This force opposes the motion as m moves a distance d = 1.5 m, so the work done by friction is
Wfric = fkd cos θ = (μkmg)(d)(−1)
Wfric = −(0.40)(3.0 kg)(9.80 m/s2)(1.5m) = −17.6J .
Mass m’s initial speed is zero, and its final speed is v. So its change in kinetic energy is
ΔK = ½ (3.0kg)v2 − 0 = (1.5kg)v2
As we noted, m has no change in potential energy during the motion. Mass M’s change in kinetic energy is
ΔK = ½ (5.0kg)v2 − 0 = (2.5kg)v2
and since it has a change in height given by −d, its change in (gravitational) potential energy is
ΔU = MgΔy = (5.0 kg)(9.80 m/s2)(−1.5m) = −73.5 J
Adding up the changes for both masses, the total change in mechanical energy of this system is
ΔE = (1.5kg)v2 + (2.5kg)v2 − 73.5 J = (4.0kg)v2 − 73.5 J
Now use ΔE = Wfric and get:
(4.0kg)v2 − 73.5J = −17.6 J
Solve for v:
(4.0kg)v2 = 55.9J
v2 = 55.9 J/4.0kg = 14.0 m2/s2
which gives
v = 3.74 m/s
The final speed of the 5.0 kg mass (in fact of both masses) is 3.74 m/s .
Problem #3
A 10.0kg block is released from point A in Fig. 4. The track is frictionless except for the portion BC, of length 6.00m. The block travels down the track, hits a spring of force constant k = 2250N/m, and compresses it 0.300m from its equilibrium position before coming to rest momentarily. Determine the coefficient of kinetic friction between surface BC and block.
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| Fig.4 |
We know that we must use energy methods to solve this problem, since the path of the sliding mass is curvy.
The forces which act on the mass as it descends and goes on to squish the spring are: gravity, the spring force and the force of kinetic friction as it slides over the rough part.
Gravity and the spring force are conservative forces, so we will keep track of them with the potential energy associated with these forces. Friction is a non-conservative force, but in this case we can calculate the work that it does. Then, we can use the energy conservation principle,
ΔK +ΔU = Wnon−cons
to find the unknown quantity in this problem, namely μk for the rough surface. We can get the answer from this equation because we have numbers for all the quantities except for Wnon−cons = Wfriction which depends on the coefficient of friction.
The block is released at point A so its initial speed (and hence, kinetic energy) is zero: Ki = 0. If we measure height upwards from the level part of the track, then the initial potential energy for the mass (all of it gravitational) is
Ui = mgh = (10.0 kg)(9.80 m/s2)(3.00 m) = 2.94 × 102 J
Next, for the “final” position of the mass, consider the time at which it has maximally compressed the spring and it is (instantaneously) at rest. (This is shown in Fig. 5.) We don’t need to think about what the mass was doing in between these two points; we don’t care about the speed of the mass during its slide.
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| Fig.5 |
Uspring = ½kx2 = ½ (2250 Nm)(0.300 m)2 = 1.01 × 102 J
so the final potential energy of the system is Uf = 1.01 × 102 J.
The total mechanical energy of the system changes because there is a non–conservative force (friction) which does work. As the mass (m) slides over the rough part, the vertical forces are gravity (mg, downward) and the upward normal force of the surface, N. As there is no vertical motion, N = mg. The magnitude of the force of kinetic friction is
fk = μkN = μkmg = μk(10.0 kg)(9.80 m/s2) = μk(98.0N)
As the block moves 6.00m this force points opposite (1800 from ) the direction of motion.
So the work done by friction is
Wfric = fkd cos θ = μk(98.0 N)(6.00 m) cos 1800 = −μk(5.88 × 102 J)
We now have everything we need to substitute into the energy balance condition, Eq. 1. We get:
(0 − 0) + (1.01 × 102 J − 2.94 × 102 J) = −μk(5.88 × 102 J)
The physics is done. We do algebra to solve for μk:
−1.93 × 102J = −μk(5.88 × 102 J)
μk = 0.328
The coefficient of kinetic friction for the rough surface and block is 0.328.
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