Question

A 0.60-kilogram softball initially at rest is hit with a bat. The ball is in contact with the bat for 0.20 second and leaves the bat with a speed of 25 meters per second. What is the magnitude of the average force exerted by the ball on the bat?
a. 15 N
b. 3.0 N
c. 75 N
d. 8.3 N

Answer 1

Answer:

c. 75 N

Explanation:

Force: This can be defined as the product of mass and the acceleration of a body. The S.I unit of Force is Newton (N).

From Newton's Fundamental equation of kinematic,

F = ma ............................................. Equation 1

Where F = Force exerted by the ball on the bat, m = mass of the ball, a = acceleration.

From motion,

a = (v-u)/t....................................... Equation 2

Where v = final velocity, u = initial velocity, t = time.

Given: v = 25 m/s, u = 0 m/s ( at rest), t = 0.20 s.

Substitute into equation 2

a = (25-0)/0.20

a = 125 m/s².

Also given: m = 0.60 kg.

Substitute into equation 1

F = 0.6(125)

F = 75 N.

Hence the force exerted by the ball on the bat = 75 N,

The right option is c. 75 N

Answer 2

The magnitude of force on the bat is required.

The force exerted buy the ball on the bat is c. 75 N.

m = Mass of softball = 0.6 kg

t = Time = 0.2 s

v = Final velocity = 25 m/s

u = Initial velocity = 0

a = Acceleration = $\dfrac{v-u}{t}$

Force is given by

$F=ma=m\dfrac{v-u}{t}\\\Rightarrow F=0.6\times \dfrac{25-0}{0.2}\\\Rightarrow F=75\ \text{N}$

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