A basketball player makes a jump shot. The 0.700-kg ball is released at a height of 2.00 m above the floor with a speed of 7.70 m/s. The ball goes through the net 3.00 m above the floor at a speed of 3.10 m/s. What is the work done on the ball by air resistance, a nonconservative force?
Calculate the loss of total mechanical energy (potential + kinetic) from release until it reaches the basket.
(1/2)M*V1^2 + M*g*H1 = (1/2)MV2^2 + M*g*H2 + (E lost)
Elost = 0.700{[(1/2)(7.7)^2-(3.1)^2] + (9.8)(2.00 - 3.00)} = ____
I consider it unlikely that a basketball would slow down that much due to air resistance, but those are the numbers you were given.
10.5 J = frictional work lost
Initial KE = 20.7 J
Something is phoney here.
To calculate the work done on the ball by air resistance, we can use the work-energy principle. The work done by a nonconservative force is equal to the change in kinetic energy of the object.
First, let's calculate the initial kinetic energy of the ball. The formula for kinetic energy is given by:
KE = (1/2) * m * v^2
where KE is the kinetic energy, m is the mass of the ball, and v is the velocity of the ball. Substituting the given values:
m = 0.700 kg
v = 7.70 m/s
KE_initial = (1/2) * 0.700 kg * (7.70 m/s)^2
Next, let's calculate the final kinetic energy of the ball. Again, using the kinetic energy formula:
m = 0.700 kg
v = 3.10 m/s
KE_final = (1/2) * 0.700 kg * (3.10 m/s)^2
Now, we can calculate the work done by air resistance using the work-energy principle:
Work = KE_final - KE_initial
Air Resistance = Work
Substitute the calculated values into the above equation to find the work done on the ball by air resistance.
Note: The net work done on the ball will be equal to the sum of work done by air resistance and the work done by other forces, like gravity.