Question

question 11 (8 points)
an air - hockey paddle hits a 48.0 g stationary puck with a force of 12.0 n. the puck travels 50 cm on the frictionless surface while the force is applied to the puck. calculate the final speed of the mass at the end of the 50 cm.

Explanation:

Step1: Convert units

First, convert the mass from grams to kilograms: \( m = 48.0\ g = 0.048\ kg \).
Convert the distance from centimeters to meters: \( d = 50\ cm = 0.50\ m \).

Step2: Calculate work done

The work done by the force is given by \( W = F \cdot d \), where \( F = 12.0\ N \) and \( d = 0.50\ m \).
So, \( W = 12.0\ N \times 0.50\ m = 6.0\ J \).

Step3: Use work - energy theorem

The work - energy theorem states that the work done on an object is equal to the change in its kinetic energy. Since the puck starts from rest, the initial kinetic energy \( KE_{i}=0 \). The final kinetic energy \( KE_{f}=\frac{1}{2}mv_{f}^{2} \), and \( W=\Delta KE = KE_{f}-KE_{i}=\frac{1}{2}mv_{f}^{2} \).
We know \( W = 6.0\ J \), \( m = 0.048\ kg \), and we can solve for \( v_{f} \):
\[

$$\begin{align*} 6.0&=\frac{1}{2}\times0.048\times v_{f}^{2}\\ v_{f}^{2}&=\frac{6.0\times2}{0.048}\\ v_{f}^{2}&=\frac{12.0}{0.048}\\ v_{f}^{2}& = 250\\ v_{f}&=\sqrt{250}\\ v_{f}&\approx15.8\ m/s \end{align*}$$

\]

Answer:

The final speed of the puck is approximately \( 15.8\ m/s \) (or more precisely \( 5\sqrt{10}\ m/s\approx15.8\ m/s \)).