3. What is the work done by the force of tension in lifting mass m a distance h?

 

    Since the tension force, T, is in the same direction as the displacement, h, we know that the work done is equal to hT. But what is the magnitude of the tension force? We know that the sum of forces acting on m is T – mg which is equal to ma. Therefore, T = m(g – a). From the solution to question 1, we know that a = g(M – m)/(M + m), so substituting in for a, we get:

    Let’s assume that mass M has already begun to slide along the table, and its movement is opposed by the force of kinetic friction, , where is the coefficient of kinetic friction, and N is the normal force acting between the mass and the table. If the mention of friction and normal forces frightens you, you might want to flip back to Chapter 3 and do a little reviewing.

 

    So let’s approach this problem with our handy three-step problem-solving method:

1. Ask yourself how the system will move: First, we know that mass m is falling and dragging mass M off the table. The force of kinetic friction opposes the motion of mass M. We also know, since both masses are connected by a nonstretching rope, that the two masses must have the same velocity and the same acceleration.
2. Choose a coordinate system: For the purposes of this problem, it will be easier if we set our coordinate system relative to the rope rather than to the table. If we say that the x-axis runs parallel to the rope, this means the x-axis will be the up-down axis for mass m and the left-right axis for mass M. Further, we can say that gravity pulls in the negative x direction. The y-axis, then, is perpendicular to the rope, and the positive y direction is away from the table.
3. Draw free-body diagrams: The above description of the coordinate system may be a bit confusing. That’s why a diagram can often be a lifesaver.

    Given this information, can you calculate the acceleration of the masses? If you think analytically and don’t panic, you can. Since they are attached by a rope, we know that both masses have the same velocity, and hence the same acceleration, a. We also know the net force acting on both masses: the net force acting on mass M is , and the net force acting on mass m is T – mg. We can then apply Newton’s Second Law to both of the masses, giving us two equations involving a:

    Adding the two equations, we find . Solving for a, we get:

    Since m is moving downward, a must be negative. Therefore, .

 

 

    It is highly unlikely that SAT II Physics will ask a question that involves remembering and then plugging numbers into an equation like this one. Remember: SAT II Physics places far less emphasis on math than your high school physics class. The test writers don’t want to test your ability to recall a formula or do some simple math. Rather, they want to determine whether you understand the formulas you’ve memorized. Here are some examples of the kinds of questions you might be asked regarding the pulley system in the free-body diagram above: