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?

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?
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Solution

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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: \[ \boxed{-252.6 \, \text{N}} \]

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