A tennis racket is in contact with a 55 kg ball for 0.0050 s changes the ball's velocity from 30 m/s [E] to 40 m/s [W] over a distance of 0.0001 m. how much work is done on the tennis ball by the racket?
Force*distance=work
but force*time=mass*changevelocity
or force= 55*(70)/.0050 Notice the 70 as change of velocity, figure that out.
One remark: I would hate to hit a 55kg ball, geepers, that is bigger than some of my high school students.
Put that force into the first work equation.
To determine the amount of work done on the tennis ball by the racket, we need to use the work-energy theorem, which states that the work done on an object is equal to the change in its kinetic energy.
The formula for work is given by:
Work = Force x Distance x cos(θ)
Where:
- Work is the work done on the object
- Force is the force applied to the object
- Distance is the displacement of the object
- θ is the angle between the applied force and the displacement
In this case, we don't have the force applied directly, but we have the change in velocity and the time over which it occurs. We can use the equation:
Work = ΔKE = (1/2) * m * (vf^2 - vi^2)
Where:
- ΔKE is the change in kinetic energy
- m is the mass of the object
- vf is the final velocity
- vi is the initial velocity
First, let's calculate the change in kinetic energy:
ΔKE = (1/2) * m * (vf^2 - vi^2)
= (1/2) * 55 kg * (40 m/s)^2 - (30 m/s)^2
Simplifying the equation:
ΔKE = (1/2) * 55 kg * (1600 m^2/s^2 - 900 m^2/s^2)
= (1/2) * 55 kg * 700 m^2/s^2
= 19250 J
Therefore, the work done on the tennis ball by the racket is 19250 Joules.