7. A 1650 kg car accelerates up a hill by applying a force of 4.61 x 10³ N parallel to the hill. The hill is at an incline of 13°. m Vo = 16 S d = 17 m 13° a. What is the net work done on the car as it travels 17 m up the hill? (Hint: calculate the work done by each force in the system. There are 3 forces. The net work is the sum of the works done by each individual force.) What is the change in kinetic energy for the car?

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### Problem 7

A \(1650 \, \text{kg}\) car accelerates up a hill by applying a force of \(4.61 \times 10^3 \, \text{N}\) parallel to the hill. The hill is at an incline of \(13^\circ\).

#### Diagram Description
The diagram shows a car moving up a hill that has an incline of \(13^\circ\). The initial speed of the car (\(v_0\)) is \(16 \, \frac{\text{m}}{\text{s}}\).
The distance (\(d\)) up the hill is \(17 \, \text{m}\).

#### Questions:
a. **What is the net work done on the car as it travels \(17 \, \text{m}\) up the hill?** 
   
   *(Hint: calculate the work done by each force in the system. There are 3 forces. The net work is the sum of the works done by each individual force.)*

b. **What is the change in kinetic energy for the car?**

c. **The car is travelling at a speed of \(16 \, \frac{\text{m}}{\text{s}}\) at the bottom of the hill. How fast is it moving at the top of the hill?**

### Solutions (Explanations)

#### Part (a) - **Net Work Done on the Car**
To calculate the net work done on the car as it travels up the hill, we need to consider the three forces acting on the car: 
1. The applied force (\(F_{\text{applied}} = 4.61 \times 10^3 \, \text{N}\)).
2. The gravitational force component along the hill.
3. The normal force, which does no work as it is perpendicular to displacement.

The net work \(W_{\text{net}}\) is given by the sum of the works done by each individual force.

#### Part (b) - **Change in Kinetic Energy**
The net work done on the car is equal to the change in kinetic energy (\(\Delta KE\)), according to the work-energy theorem:
\[ \Delta KE = W_{\text{net}} \]

#### Part (c) - **Speed at the Top of the Hill**
To find the speed at the top of the hill, we use the final kinetic energy:
\[ KE_{\text{
Transcribed Image Text:### Problem 7 A \(1650 \, \text{kg}\) car accelerates up a hill by applying a force of \(4.61 \times 10^3 \, \text{N}\) parallel to the hill. The hill is at an incline of \(13^\circ\). #### Diagram Description The diagram shows a car moving up a hill that has an incline of \(13^\circ\). The initial speed of the car (\(v_0\)) is \(16 \, \frac{\text{m}}{\text{s}}\). The distance (\(d\)) up the hill is \(17 \, \text{m}\). #### Questions: a. **What is the net work done on the car as it travels \(17 \, \text{m}\) up the hill?** *(Hint: calculate the work done by each force in the system. There are 3 forces. The net work is the sum of the works done by each individual force.)* b. **What is the change in kinetic energy for the car?** c. **The car is travelling at a speed of \(16 \, \frac{\text{m}}{\text{s}}\) at the bottom of the hill. How fast is it moving at the top of the hill?** ### Solutions (Explanations) #### Part (a) - **Net Work Done on the Car** To calculate the net work done on the car as it travels up the hill, we need to consider the three forces acting on the car: 1. The applied force (\(F_{\text{applied}} = 4.61 \times 10^3 \, \text{N}\)). 2. The gravitational force component along the hill. 3. The normal force, which does no work as it is perpendicular to displacement. The net work \(W_{\text{net}}\) is given by the sum of the works done by each individual force. #### Part (b) - **Change in Kinetic Energy** The net work done on the car is equal to the change in kinetic energy (\(\Delta KE\)), according to the work-energy theorem: \[ \Delta KE = W_{\text{net}} \] #### Part (c) - **Speed at the Top of the Hill** To find the speed at the top of the hill, we use the final kinetic energy: \[ KE_{\text{
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