A catcher picks up a baseball from the ground. If the net upward force on the ball is 7.25 × 10–2 N and the net

work done lifting the ball is 4.35 × 10–2 J, how far is the ball lifted?

1 answer

To find the distance the ball is lifted, we can use the formula for work done, which is given by:

\[ W = F \cdot d \cdot \cos(\theta) \]

where:

  • \( W \) is the work done (in joules),
  • \( F \) is the net force (in newtons),
  • \( d \) is the distance (in meters),
  • \( \theta \) is the angle between the force and the direction of motion.

In this case, since the ball is being lifted straight upwards, the angle \( \theta = 0^\circ \) and \( \cos(0^\circ) = 1\). Therefore, the equation simplifies to:

\[ W = F \cdot d \]

Rearranging this to solve for \( d \):

\[ d = \frac{W}{F} \]

Now substitute the given values into the equation:

\[ W = 4.35 \times 10^{-2} , \text{J} \] \[ F = 7.25 \times 10^{-2} , \text{N} \]

Now plug in these values:

\[ d = \frac{4.35 \times 10^{-2}}{7.25 \times 10^{-2}} \]

Calculating this gives:

\[ d = \frac{4.35}{7.25} \times \frac{10^{-2}}{10^{-2}} = \frac{4.35}{7.25} \approx 0.6 \]

Therefore, the distance the ball is lifted is approximately:

\[ d \approx 0.60 , \text{m} \]

This means the ball is lifted approximately 0.60 meters.