Question

A pitcher throws a 0.140 kg baseball, and it approaches the bat at a speed of 35.0 m/s. The bat does Wnc = 75.0 J of work on the ball in hitting it. Ignoring air resistance, determine the speed of the ball after the ball leaves the bat and is 25.0 m above the point of impact.(Answer in m/s)

Answer 1

Concepts and reason

The concepts used to solve this problem are non-conservative forces and work energy theorem.

First, use the kinetic and potential energy expressions to find the expression for non-conservative force in terms of initial and final velocity of kinetic energy and initial and final heights of potential energy.

Then, using the expression for work done, find the work done by non-conservative force.

Fundamentals

“The work done by non-conservative force is equal to the change in mechanical energy”.

The expression for the work done by non-conservative force is,

Here, work done by non-conservative force is , final kinetic energy is , initial kinetic energy is , final potential energy is , and initial potential energy is .

The energy possesses due to the motion object or particle is called kinetic energy.

The expression for kinetic energy is,

Here, kinetic energy is , mass of the object is , and velocity is .

The saved energy in an object due to its position is called potential energy.

The expression for potential energy is,

Here, potential energy is , gravitational constant is , and height is .

The expression for kinetic energy is,

Similarly, the expression for initial kinetic energy is,

…… (1)

Here, initial velocity is .

The expression for final kinetic energy is,

…… (2)

Here, final velocity is .

The expression for potential energy is,

Similarly, the expression for initial potential energy is,

…… (3)

Here, initial height is .

The expression for final potential energy is,

…… (4)

Here, final height is .

The expression for the work done by non-conservative force is,

Substitute equation (1), (2), (3), and (4) in the above equation,

The initial height of the ball leaves the bat is zero (i.e.) so, the expression becomes,

…… (5)

From equation (5),

Substitute for , for , for , for , and for .

Rearrange the equation in terms of .

Taking square root on both the sides,

Ans:

The final speed of the ball after it leaves the bat and is above the point of impact is .