Questions: You have a 52 kg object traveling at 17 m / s. That object hits a rough patch of ground and slides to a stop in 3.5 seconds. What Force does friction provide to this object?
Transcript text: You have a 52 kg object traveling at $17 \mathrm{~m} / \mathrm{s}$. That object hits a rough patch of ground and slides to a stop in 3.5 seconds. What Force does friction provide to this object?
Solution
Solution Steps
Step 1: Determine the Initial Momentum of the Object
The initial momentum \( p \) of the object can be calculated using the formula:
\[
p = m \cdot v
\]
where \( m = 52 \, \text{kg} \) is the mass of the object and \( v = 17 \, \text{m/s} \) is its initial velocity. Thus,
\[
p = 52 \, \text{kg} \times 17 \, \text{m/s} = 884 \, \text{kg} \cdot \text{m/s}
\]
Step 2: Calculate the Change in Momentum
Since the object comes to a stop, its final velocity is \( 0 \, \text{m/s} \). Therefore, the final momentum is:
\[
p_{\text{final}} = 52 \, \text{kg} \times 0 \, \text{m/s} = 0 \, \text{kg} \cdot \text{m/s}
\]
The change in momentum \( \Delta p \) is:
\[
\Delta p = p_{\text{final}} - p_{\text{initial}} = 0 - 884 = -884 \, \text{kg} \cdot \text{m/s}
\]
Step 3: Use Impulse-Momentum Theorem to Find the Force
The impulse-momentum theorem states that the change in momentum is equal to the impulse applied to the object, which is the product of the force and the time duration:
\[
\Delta p = F \cdot \Delta t
\]
where \( \Delta t = 3.5 \, \text{s} \) is the time duration. Solving for the force \( F \), we have:
\[
F = \frac{\Delta p}{\Delta t} = \frac{-884 \, \text{kg} \cdot \text{m/s}}{3.5 \, \text{s}} = -252.5714 \, \text{N}
\]
Final Answer
The force of friction acting on the object is:
\[
\boxed{-252.6 \, \text{N}}
\]