Energy and Momentum Test Review Explanations


 Adrian Hampton
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1 1. Zero net work on an object means that no net energy is being transferred into or out of the object, so the object is not changing its speed. In the given graph, the object's speed is only changing during the interval from 2 to 3 seconds, seen by the changing slope of the graph during that interval. So if the object is moving at a constant speed during 0 to 1 seconds, and not moving at all from 1 to 2 seconds, there must also be zero net work during those two intervals. 2. First, it is important to recognize that the question is asking for power, because power is defined as the rate at which work is done. Power is calculated by P = W / t, but work is not given in this question. So to begin, work must be calculated using the WorkKinetic Energy Theorem. Then to finish solving, work must be divided by 4 seconds of time, to arrive at 110 W. 3. Positive net work occurs when energy is transferred INTO an object, leading to an increase in the object's speed. So positive net work is occurring anytime an object speeds up. On the given graph, the object is slowing down from 0 to 1 second, as shown by the velocity getting closer and closer to zero. From 1 to 2 seconds on the graph, the object's velocity is getting farther and farther from zero, meaning that the object is speeding up (in the negative direction). From 2 to 3 seconds, the object is again slowing down, because its velocity is getting closer and closer to zero. Notice that positive work really has nothing to do with whether the object is traveling in the positive direction. Work is a scalar measurement, not a vector. Positive work doesn't indicate direction, but instead indicates the flow of energy into an object. 4. Work can be calculated as the "area under the curve" on a force versus position graph. The simplest way to calculate this total area is to view it as a triangular area to the left of a rectangular area. These two areas, and then the total work, can be found by But this question asks for final speed, not for work, so the WorkKinetic Energy Theorem must be employed.
2 5. Work done by gravity is found by multiplying gravitational force (or "weight") by the component of displacement parallel to the gravitational force. The easiest way to think about it on this problem is to simply multiply weight by the VERTICAL distance moved by the crate. 6. Final speed can be calculated using the WorkKinetic Energy Theorem. The only slightly complicated aspect is that the force doing work pulls at an angle above the horizontal, so the angle must be factored into the work calculation. 7. Because friction acts during part of the motion, the mechanical energy of the box does not remain constant. But the mechanical energy changes by an amount exactly equal to the nonconservative work done by friction. So final ME can be set equal to initial ME plus the negative work done by friction. At the beginning of the motion, the box possesses only kinetic energy. When the box reaches point P it possesses both kinetic energy and potential energy (with the initial height as the reference level). The only other part of the equation to consider is the work done by friction, which is found with frictional force, rough distance, and cosine 180. The equation and solution are as follows. 8. The total mechanical energy of the mass and spring system will remain constant throughout the motion, so initial ME can be set equal to final ME. When the ball and spring are first released, the system possesses elastic potential energy (because the spring is compressed) but no gravitational potential energy (if this lowest position is viewed as the reference level) and no kinetic energy (because there is no motion yet). When the ball is passing through equilibrium, there is no elastic potential energy (because the spring is at equilibrium) but non
3 zero gravitational potential energy (because the ball has moved upwards) and nonzero kinetic energy (because the ball is moving). So the problem can be solved by: 9. Because friction is acting, the mechanical energy of the box does not remain constant throughout this motion. But the mechanical energy changes by an amount exactly equal to the nonconservative work done by friction. So final ME can be set equal to initial ME plus the negative work done by friction. Calculating the work done by friction is the hardest part of this problem. First the frictional force must be determined with the coefficient of friction and the normal force, which is equal to the component of gravity perpendicular to the ramp. Work by friction also requires the distance the box moves along the incline, which can be related to final height H using sine. The angle in the work formula is always the angle between the force and distance, so in this case the angle is 180 and the cosine gives 1. Now it is relatively simple to set up the final energyconservation equation. At the base of the incline, the box possesses only kinetic energy, and when it reaches its final height it possesses only gravitational potential energy. So the equation, including work done by friction, becomes 10. The given type of graph can be used to find impulse (F t), by finding the area under the curve.
4 Once impulse is known, it can be used to calculate the object's final speed. (It is important to also note that this object did not start from rest, but instead had a speed of 4 m/s at time t=0.) 11. Momentum is conserved in all types of collisions, so the system's total momentum before the collision must be equal to the system's total momentum after the collision. The application of momentum conservation ideas, then, is as follows: 12. Because the mother and child start at rest, the system's total initial momentum is zero. And because momentum must be conserved for the motherchild system, the system must still have zero total momentum later after they push off. The only way this can happen is if the mother and child both end with the same amount of momentum, in opposite directions. Because they have different masses, they do not have the same speed after pushing. And if momentum (mv) is equal, then kinetic energy (½mv 2 ) will generally not be equal.
5 13. During the sliding part of the problem, there is no friction so the mechanical energy of the 2 kg block is conserved. So energyconservation ideas can be used to calculate the speed of the block at the end of its slide, before the collision. During the collision, momentum of the entire system is conserved, so momentumconservation ideas can be used to solve for final speed.
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