All questions of Work and Energy of Point Object Systems (PHY) for MCAT Exam

The work–energy theorem relates the total work done on an object by all forces to the change in kinetic energy experienced by the same object. While the work done by a force is indeed proportional to the magnitude of the force, it is also proportional to the displacement of the object, eliminating choice (A). The change in kinetic energy is equal—not proportional—to the total work done on the object; further, it is the net force, not any force, that relates to the work done on an object, eliminating choice (C). Finally, the change in kinetic energy of the object is equal to the work done by all of the forces acting on the object combined, not just the applied force, which eliminates choice (D).

Sarah will not bounce higher than Josh. Assuming that mechanical energy is conserved, Sarah and Josh will start with a given amount of potential energy, which is converted into kinetic energy, then elastic potential energy, then kinetic energy again with no loss of energy from the system, eliminating choice (D). By this logic, both individuals should return to the same starting height. Josh starts with

of potential energy. At the moment he hits the net, all of this potential energy has been converted into kinetic energy. Therefore,

eliminating choice (B). Josh will experience a greater force upon impact because the net exerts a force proportional to weight; the higher the weight, the larger the force exerted by the net, eliminating choice (C).

Mechanical advantage is a ratio of the output force generated given a particular input force. Efficiency is a ratio of the useful work performed by a system compared to the work performed on the system.

Because the weight of the barbell (force acting downward) is or about 2750 N, it follows that the weightlifter must exert an equal and opposite force of 2750 N on the barbell. The work done in lifting the barbell is therefore W = Fd cos θ = (2750 N)(2.4 m)(cos 0) ≈ 7000 J. Using the same equation, it follows that the work done to hold the barbell in place is W = Fd cos θ = (2750 N)(0 m)(cos θ) = 0 J. Because the barbell is held in place and there is no displacement, the work done is zero.

Gravity is a conservative force because it is pathway independent and it does not dissipate mechanical energy. Air resistance and friction—choices (A) and (B)—are nonconservative forces that dissipate energy thermally. Convection is not a force, but a method of heat transfer, eliminating choice (D).

An isochoric process, by definition, is one in which the gas system undergoes no change in volume. If the gas neither expands nor is compressed, then no work is performed. Remember that work in a thermodynamic system is the area under a P–V curve; if the change in volume is 0, then the area under the curve is also 0.

The work done by the tractor can be calculated from the equation
W = Fd cos θ = (5000 N)(100 m)(cos 30°) = (5000)(100)(0.866) ≈ 5000 × 80 = 400,000 J = 400 kJ.

The work done by the engine is equal to the change in kinetic energy of the car:

The average power therefore is

Horsepower is a unit of power, as evidenced by the name and the conversion factor given in the question stem. Power is a rate of energy expenditure over time. Given unlimited time, both cars are capable of unlimited increases in (kinetic) energy, meaning that they have unlimited maximum velocities. The fact that Car B has a higher power rating means that it will reach any given velocity faster than Car A, eliminating choice (B). There is not enough information to make any judgments on the efficiency of the cars, eliminating choice (C). While it may take longer for Car A to reach a given velocity, both cars have unlimited maximum velocities according to the information given in the stem, eliminating choice (D).

Assuming negligible air resistance, conservation of energy states that the total mechanical energy of the block is constant as it falls. At the starting height of 5 m, the block only has potential energy equal to

Because the kinetic energy at this point is 0 J, the total mechanical energy is 2000 J at any point during the block's descent.