Work and Power

The velocity of the ball is easily calculable: v = = 10 m/s . With values for the mass and velocity of the ball, we can calculate kinetic energy:

To find the velocity of the man we must find how far he travels over a given time period. In one day, or 86400 seconds, the man travels the circumference of the earth, or 2Πr meters. Thus the velocity of the man is v = = = 463 m/s. Again, since we know the velocity and mass of the man we can calculate kinetic energy. K = mv ² = (50 kg)(463 m/s)² = 5.36×10⁶ Joules.

The ball is acted upon by a constant gravitational force, mg . The work done during its total trip, then, is simply mgh . By the Work-Energy theorem, this causes a change in kinetic energy. Since the ball initially has no velocity, we can find the final velocity by the equation:

Solving for v ,

The final velocity of the ball is 14 m/s. We found this by one simple calculation, avoiding the cumbersome kinematic equations. This is an excellent demonstration of the advantages of working with the concepts of work and energy, as opposed to simple kinematics.

The ball reaches its maximum height when its velocity is reduced to zero. This change in velocity is caused by the work done by gravitational force. We know the change in velocity, and hence the change in kinetic energy of the ball, and can calculate its maximum height from this:

But v _f = 0 , and the masses cancel, so

When the ball is at a height of 25 meters, the gravitational force has done an amount of work on the ball equal to W = - mgh = - 25 mg. This work causes a change in velocity of the particle. We now use the Work-Energy Theorem, and solve for the final velocity:

Again, the masses cancel:

Thus

At the top of the loop, the ball must have enough velocity such that the centripetal force provided by its weight keeps the ball in circular motion. In other words:

Solving for v ,

This value for the velocity gives us the minimum velocity at the top of the loop. But we are asked for the minimum velocity at the bottom of the loop. How do we find this? You guessed it: Work-Energy Theorem.

During the entire vertical loop, the ball is acted upon by two forces: the normal force and the gravitational force. The normal force, by definition, always points perpendicular to the circumference of the loop, and thus the motion of the ball. Consequently, it cannot perform work on the ball. The gravitational force, on the other hand, does perform work on the ball, according the height it reaches. Since the radius of the circle is 2 m, the ball reaches a height of 4 m, and experiences work from the gravitational force of - mgh = - 2mg . Remember the sign is negative because the force acts in a direction opposite the motion of the ball. This work causes a change in velocity from the bottom of the loop to the top of the loop, which can be calculated by the work-energy theorem:

Thus

Canceling the mass and solving for v _o ,

Thus the ball must enter the vertical loop of at least 7.7 m/s.