Question

1. a baseball player exerts a force of 100 n on a ball for a distance of 0.5 m as he throws it. if the ball has a mass of 0.15 kg, what is its velocity as it leaves his hand?

a. -36.5 m/s
b. -25.8 m/s
c. 25.8 m/s
d. 36.5 m/s

Explanation:

Step1: Calculate Work Done

Work \( W \) is force \( F \) times distance \( d \), so \( W = F \times d \).
Given \( F = 100\,\text{N} \), \( d = 0.5\,\text{m} \),
\( W = 100 \times 0.5 = 50\,\text{J} \).

Step2: Relate Work to Kinetic Energy

Work done equals kinetic energy \( KE = \frac{1}{2}mv^2 \).
Mass \( m = 0.15\,\text{kg} \), \( KE = 50\,\text{J} \),
\( 50 = \frac{1}{2} \times 0.15 \times v^2 \).

Step3: Solve for Velocity \( v \)

Rearrange: \( v^2 = \frac{2 \times 50}{0.15} = \frac{100}{0.15} \approx 666.67 \).
Take square root: \( v = \sqrt{666.67} \approx 25.8\,\text{m/s} \) (velocity is positive as it's magnitude here).

Answer:

c. 25.8 m/s