Question

A 0.140 kg baseball is thrown horizontally with a velocity of 28.9 m/s. It is struck with a constant horizontal force that lasts for 1.85 ms, which results in a velocity of 37.0 m/s in the opposite direction. What was the magnitude of the horizontal force? a. 0.613 kNb. 0.613 kN c. 2.19 kN d. 2.80 kN

Answer 1

To solve this problem, we will express the Force according to Newton's second Law, and acceleration, such as the change of velocity in the body by a certain distance, therefore,

The magnitude of the force is given as

F = ma

Where,

m = mass

a = Acceleration

That can be expressed as,

`F = m(dv)/(dt)`

Replacing with our values we have

F = 0.140kg * (\frac{37+28.9}{1.85*10^{-3}})

F = 4987.03N

F= 4.9kN

Therefore the magnitude of the horizontal force is 4.9kN