A sled loaded with firewood is pulled by a tractor. The tractor exerts a constant force with magnitude 2000 N in the horizontal direction. The total weight of the sled and the firewood is 4000 N, and the coefficient of kinetic friction between the sled and the road is 0.200. If the sled travels a distance of 28.5 m, calculate the final velocity of the sled using work-energy theorem (its initial velocity is zero). Maintain standard units. Normal format and 3 SF.

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**Question:**

A sled loaded with firewood is pulled by a tractor. The tractor exerts a constant force with a magnitude of 2000 N in the horizontal direction. The total weight of the sled and the firewood is 4000 N, and the coefficient of kinetic friction between the sled and the road is 0.200. If the sled travels a distance of 28.5 m, calculate the final velocity of the sled using the work-energy theorem (its initial velocity is zero). Maintain standard units. Normal format and 3 SF.

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This question requires applying the work-energy theorem to determine the sled's final velocity, given specific forces and conditions. Here’s a breakdown of how to approach the solution:

1. **Determine the work done by the tractor:**
   - Work done by the tractor \( W = \text{Force} \times \text{Distance} = 2000 \, \text{N} \times 28.5 \, \text{m} \).

2. **Calculate the frictional force:**
   - Frictional force \( f = \text{coefficient of friction} \times \text{normal force} = 0.200 \times 4000 \, \text{N} \).

3. **Calculate the work done against friction:**
   - Work against friction \( W_{\text{friction}} = f \times \text{Distance} = \text{frictional force} \times 28.5 \, \text{m} \).

4. **Apply the work-energy theorem:**
   - Net work \( W_{\text{net}} = W - W_{\text{friction}} \).
   - Use this to find the final kinetic energy: 
     \[
     W_{\text{net}} = \frac{1}{2}mv^2 - \frac{1}{2}m(0)^2
     \]
   - Solve for \( v \) (final velocity).

Remember to convert the weight to mass by using \( W = mg \) if needed, where \( g = 9.81 \, \text{m/s}^2 \). Make sure the units are consistent throughout the calculation.
Transcribed Image Text:**Question:** A sled loaded with firewood is pulled by a tractor. The tractor exerts a constant force with a magnitude of 2000 N in the horizontal direction. The total weight of the sled and the firewood is 4000 N, and the coefficient of kinetic friction between the sled and the road is 0.200. If the sled travels a distance of 28.5 m, calculate the final velocity of the sled using the work-energy theorem (its initial velocity is zero). Maintain standard units. Normal format and 3 SF. --- This question requires applying the work-energy theorem to determine the sled's final velocity, given specific forces and conditions. Here’s a breakdown of how to approach the solution: 1. **Determine the work done by the tractor:** - Work done by the tractor \( W = \text{Force} \times \text{Distance} = 2000 \, \text{N} \times 28.5 \, \text{m} \). 2. **Calculate the frictional force:** - Frictional force \( f = \text{coefficient of friction} \times \text{normal force} = 0.200 \times 4000 \, \text{N} \). 3. **Calculate the work done against friction:** - Work against friction \( W_{\text{friction}} = f \times \text{Distance} = \text{frictional force} \times 28.5 \, \text{m} \). 4. **Apply the work-energy theorem:** - Net work \( W_{\text{net}} = W - W_{\text{friction}} \). - Use this to find the final kinetic energy: \[ W_{\text{net}} = \frac{1}{2}mv^2 - \frac{1}{2}m(0)^2 \] - Solve for \( v \) (final velocity). Remember to convert the weight to mass by using \( W = mg \) if needed, where \( g = 9.81 \, \text{m/s}^2 \). Make sure the units are consistent throughout the calculation.
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