Question
A 12-4g block is pushed 20 m up the sloping surface of a plane inclined at an angle of 37 to the horizontal, by a constant horizontal P of 240 N. The coefficient of friction between the block and plane
0.25
What is the work of the force P7
b What is the work done by the friction furce?
What is the work done by gravity?
What is the total work done on the block?
0.25
What is the work of the force P7
b What is the work done by the friction furce?
What is the work done by gravity?
What is the total work done on the block?
Answers
GPT-4o mini
To solve the problem, let's start with the basic principles of work, forces, and gravity.
Given data:
- Mass of the block, \( m = 12.4 \) kg
- Distance moved up the plane, \( d = 20 \) m
- Angle of the incline, \( \theta = 37^\circ \)
- Constant horizontal force, \( P = 240 \) N
- Coefficient of friction, \( \mu = 0.25 \)
### a) Work done by the force \( P \)
Since the force \( P \) is horizontal, we need to find the component of this force that is parallel to the displacement on the incline. The incline forms an angle of \( \theta = 37^\circ \) with the horizontal.
To find the component of force \( P \) acting up the incline, we first find the angle \( \alpha \) associated with the incline to the horizontal components:
The angle \( \alpha \) (with respect to the incline) can be found by determining the angle of incline:
\[
\text{Vertical component of } P = P \sin(\theta)
\]
\[
\text{Horizontal component of } P = P \cos(\theta)
\]
Since \( P \) is not acting directly along the incline, we will convert this into an effective force along the incline.
Using trigonometry:
The effective component of the force \( P \) along the incline:
\[
\text{Work done by } P = P_{\text{incline}} \times d = P \cos(\theta) \times d
\]
Calculating:
\[
\theta = 37^\circ
\]
\[
\cos(37^\circ) \approx 0.798
\]
\[
\text{Work done by } P = 240 \, \text{N} \times 20 \, \text{m} \times \cos(37^\circ) = 240 \times 20 \times 0.798 \approx 4785.6 \, \text{J}
\]
### b) Work done by the friction force
To find the frictional force, we first determine the normal force \( N \) acting on the block. The normal force can be calculated using the weight of the block and its components due to the incline:
1. Calculate weight (force of gravity):
\[
W = mg = 12.4 \text{ kg} \times 9.81 \text{ m/s}^2 \approx 121.5 \text{ N}
\]
The normal force is affected by the angle:
\[
N = W \cos(37^\circ) = 121.5 \text{ N} \times \cos(37^\circ) \approx 121.5 \times 0.798 \approx 96.9 \text{ N}
\]
2. Calculate the frictional force:
\[
F_{\text{friction}} = \mu N = 0.25 \times 96.9 \approx 24.2 \text{ N}
\]
3. Now, since friction acts in the opposite direction to the motion up the incline, the work done by friction will be negative:
\[
W_{\text{friction}} = -F_{\text{friction}} \times d = -24.2 \, \text{N} \times 20 \, \text{m} = -484 \, \text{J}
\]
### c) Work done by gravity
The work done by gravity when moving up the incline can be determined by calculating the component of the weight acting down the incline:
\[
W_{\text{gravity}} = -W \sin(37^\circ) \times d
\]
Calculating:
\[
\sin(37^\circ) \approx 0.601
\]
\[
W_{\text{gravity}} = -121.5 \text{ N} \times \sin(37^\circ) \times d
\]
\[
W_{\text{gravity}} = -121.5 \text{ N} \times 0.601 \times 20 \approx -145.3 \text{ J}
\]
### d) Total work done on the block
The total work done on the block can be calculated by summing up the work done by all forces (work by \( P \), work by friction, and work by gravity):
\[
W_{\text{total}} = W_{P} + W_{\text{friction}} + W_{\text{gravity}}
\]
Calculating:
\[
W_{\text{total}} = 4785.6 \, \text{J} - 484 \, \text{J} - 145.3 \, \text{J} \approx 4156.3 \, \text{J}
\]
### Final Answers:
- **Work of the force \( P \)**: Approximately **4785.6 J**
- **Work done by the friction force**: Approximately **-484 J**
- **Work done by gravity**: Approximately **-145.3 J**
- **Total work done on the block**: Approximately **4156.3 J**
Given data:
- Mass of the block, \( m = 12.4 \) kg
- Distance moved up the plane, \( d = 20 \) m
- Angle of the incline, \( \theta = 37^\circ \)
- Constant horizontal force, \( P = 240 \) N
- Coefficient of friction, \( \mu = 0.25 \)
### a) Work done by the force \( P \)
Since the force \( P \) is horizontal, we need to find the component of this force that is parallel to the displacement on the incline. The incline forms an angle of \( \theta = 37^\circ \) with the horizontal.
To find the component of force \( P \) acting up the incline, we first find the angle \( \alpha \) associated with the incline to the horizontal components:
The angle \( \alpha \) (with respect to the incline) can be found by determining the angle of incline:
\[
\text{Vertical component of } P = P \sin(\theta)
\]
\[
\text{Horizontal component of } P = P \cos(\theta)
\]
Since \( P \) is not acting directly along the incline, we will convert this into an effective force along the incline.
Using trigonometry:
The effective component of the force \( P \) along the incline:
\[
\text{Work done by } P = P_{\text{incline}} \times d = P \cos(\theta) \times d
\]
Calculating:
\[
\theta = 37^\circ
\]
\[
\cos(37^\circ) \approx 0.798
\]
\[
\text{Work done by } P = 240 \, \text{N} \times 20 \, \text{m} \times \cos(37^\circ) = 240 \times 20 \times 0.798 \approx 4785.6 \, \text{J}
\]
### b) Work done by the friction force
To find the frictional force, we first determine the normal force \( N \) acting on the block. The normal force can be calculated using the weight of the block and its components due to the incline:
1. Calculate weight (force of gravity):
\[
W = mg = 12.4 \text{ kg} \times 9.81 \text{ m/s}^2 \approx 121.5 \text{ N}
\]
The normal force is affected by the angle:
\[
N = W \cos(37^\circ) = 121.5 \text{ N} \times \cos(37^\circ) \approx 121.5 \times 0.798 \approx 96.9 \text{ N}
\]
2. Calculate the frictional force:
\[
F_{\text{friction}} = \mu N = 0.25 \times 96.9 \approx 24.2 \text{ N}
\]
3. Now, since friction acts in the opposite direction to the motion up the incline, the work done by friction will be negative:
\[
W_{\text{friction}} = -F_{\text{friction}} \times d = -24.2 \, \text{N} \times 20 \, \text{m} = -484 \, \text{J}
\]
### c) Work done by gravity
The work done by gravity when moving up the incline can be determined by calculating the component of the weight acting down the incline:
\[
W_{\text{gravity}} = -W \sin(37^\circ) \times d
\]
Calculating:
\[
\sin(37^\circ) \approx 0.601
\]
\[
W_{\text{gravity}} = -121.5 \text{ N} \times \sin(37^\circ) \times d
\]
\[
W_{\text{gravity}} = -121.5 \text{ N} \times 0.601 \times 20 \approx -145.3 \text{ J}
\]
### d) Total work done on the block
The total work done on the block can be calculated by summing up the work done by all forces (work by \( P \), work by friction, and work by gravity):
\[
W_{\text{total}} = W_{P} + W_{\text{friction}} + W_{\text{gravity}}
\]
Calculating:
\[
W_{\text{total}} = 4785.6 \, \text{J} - 484 \, \text{J} - 145.3 \, \text{J} \approx 4156.3 \, \text{J}
\]
### Final Answers:
- **Work of the force \( P \)**: Approximately **4785.6 J**
- **Work done by the friction force**: Approximately **-484 J**
- **Work done by gravity**: Approximately **-145.3 J**
- **Total work done on the block**: Approximately **4156.3 J**