Question
A 12kg 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 P?
b What is the work done by the friction force?
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 P?
b What is the work done by the friction force?
What is the work done by gravity?
What is the total work done on the block?
Answers
GPT-4o mini
To calculate the work done by various forces on the 12-kg block, we need to follow these steps:
### Given Data
- Mass of the block, \( m = 12 \, \text{kg} \)
- Distance moved along the incline, \( d = 20 \, \text{m} \)
- Angle of incline, \( \theta = 37^\circ \)
- Horizontal force \( P = 240 \, \text{N} \)
- Coefficient of friction \( \mu = 0.25 \)
### a. Work done by the force \( P \)
The work done by a force is calculated using the formula:
\[
W = F \cdot d \cdot \cos(\phi)
\]
where \( \phi \) is the angle between the force and the direction of motion. Here, since \( P \) is horizontal and the block moves up the incline, we find \( \phi \) as follows:
\[
\phi = 90^\circ - \theta
\]
\[
\phi = 90^\circ - 37^\circ = 53^\circ
\]
Now we can calculate the work done by force \( P \):
\[
W_P = P \cdot d \cdot \cos(53^\circ) = 240 \, \text{N} \cdot 20 \, \text{m} \cdot \cos(53^\circ)
\]
Using \(\cos(53^\circ) \approx 0.6018\):
\[
W_P = 240 \cdot 20 \cdot 0.6018 \approx 2889.12 \, \text{J}
\]
### b. Work done by the friction force
The frictional force \( f \) is calculated as follows:
\[
f = \mu \cdot N
\]
Here, \( N \) is the normal force. The normal force on an incline is given by:
\[
N = mg \cos(\theta)
\]
First, calculate \( mg \):
\[
mg = 12 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 \approx 117.72 \, \text{N}
\]
Now calculate the normal force \( N \):
\[
N = 117.72 \cos(37^\circ) \approx 117.72 \cdot 0.7986 \approx 93.83 \, \text{N}
\]
Now calculate the frictional force \( f \):
\[
f = \mu \cdot N = 0.25 \cdot 93.83 \approx 23.46 \, \text{N}
\]
The work done by the friction force (which opposes the movement) is:
\[
W_{\text{friction}} = - f \cdot d = -23.46 \, \text{N} \cdot 20 \, \text{m} \approx -469.2 \, \text{J}
\]
### c. Work done by gravity
The work done by gravity is calculated as follows:
The component of the weight of the block acting along the incline is:
\[
W_{\text{gravity}} = - m g \sin(\theta) \cdot d
\]
Calculating \( m g \):
\[
W_{\text{gravity}} = -12 \cdot 9.81 \cdot \sin(37^\circ) \cdot 20
\]
Calculating \( \sin(37^\circ) \approx 0.6018 \):
\[
W_{\text{gravity}} = -12 \cdot 9.81 \cdot 0.6018 \cdot 20
\]
\[
W_{\text{gravity}} \approx -12 \cdot 9.81 \cdot 0.6018 \cdot 20 \approx -141.23 \, \text{J}
\]
### d. Total work done on the block
Now we can find the total work done on the block by summing:
\[
W_{\text{total}} = W_P + W_{\text{friction}} + W_{\text{gravity}}
\]
\[
W_{\text{total}} \approx 2889.12 - 469.2 - 141.23
\]
\[
W_{\text{total}} \approx 2278.69 \, \text{J}
\]
### Summary
1. Work done by the force \( P \): \( \approx 2889.12 \, \text{J} \)
2. Work done by the friction force: \( \approx -469.2 \, \text{J} \)
3. Work done by gravity: \( \approx -141.23 \, \text{J} \)
4. Total work done on the block: \( \approx 2278.69 \, \text{J} \)
### Given Data
- Mass of the block, \( m = 12 \, \text{kg} \)
- Distance moved along the incline, \( d = 20 \, \text{m} \)
- Angle of incline, \( \theta = 37^\circ \)
- Horizontal force \( P = 240 \, \text{N} \)
- Coefficient of friction \( \mu = 0.25 \)
### a. Work done by the force \( P \)
The work done by a force is calculated using the formula:
\[
W = F \cdot d \cdot \cos(\phi)
\]
where \( \phi \) is the angle between the force and the direction of motion. Here, since \( P \) is horizontal and the block moves up the incline, we find \( \phi \) as follows:
\[
\phi = 90^\circ - \theta
\]
\[
\phi = 90^\circ - 37^\circ = 53^\circ
\]
Now we can calculate the work done by force \( P \):
\[
W_P = P \cdot d \cdot \cos(53^\circ) = 240 \, \text{N} \cdot 20 \, \text{m} \cdot \cos(53^\circ)
\]
Using \(\cos(53^\circ) \approx 0.6018\):
\[
W_P = 240 \cdot 20 \cdot 0.6018 \approx 2889.12 \, \text{J}
\]
### b. Work done by the friction force
The frictional force \( f \) is calculated as follows:
\[
f = \mu \cdot N
\]
Here, \( N \) is the normal force. The normal force on an incline is given by:
\[
N = mg \cos(\theta)
\]
First, calculate \( mg \):
\[
mg = 12 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 \approx 117.72 \, \text{N}
\]
Now calculate the normal force \( N \):
\[
N = 117.72 \cos(37^\circ) \approx 117.72 \cdot 0.7986 \approx 93.83 \, \text{N}
\]
Now calculate the frictional force \( f \):
\[
f = \mu \cdot N = 0.25 \cdot 93.83 \approx 23.46 \, \text{N}
\]
The work done by the friction force (which opposes the movement) is:
\[
W_{\text{friction}} = - f \cdot d = -23.46 \, \text{N} \cdot 20 \, \text{m} \approx -469.2 \, \text{J}
\]
### c. Work done by gravity
The work done by gravity is calculated as follows:
The component of the weight of the block acting along the incline is:
\[
W_{\text{gravity}} = - m g \sin(\theta) \cdot d
\]
Calculating \( m g \):
\[
W_{\text{gravity}} = -12 \cdot 9.81 \cdot \sin(37^\circ) \cdot 20
\]
Calculating \( \sin(37^\circ) \approx 0.6018 \):
\[
W_{\text{gravity}} = -12 \cdot 9.81 \cdot 0.6018 \cdot 20
\]
\[
W_{\text{gravity}} \approx -12 \cdot 9.81 \cdot 0.6018 \cdot 20 \approx -141.23 \, \text{J}
\]
### d. Total work done on the block
Now we can find the total work done on the block by summing:
\[
W_{\text{total}} = W_P + W_{\text{friction}} + W_{\text{gravity}}
\]
\[
W_{\text{total}} \approx 2889.12 - 469.2 - 141.23
\]
\[
W_{\text{total}} \approx 2278.69 \, \text{J}
\]
### Summary
1. Work done by the force \( P \): \( \approx 2889.12 \, \text{J} \)
2. Work done by the friction force: \( \approx -469.2 \, \text{J} \)
3. Work done by gravity: \( \approx -141.23 \, \text{J} \)
4. Total work done on the block: \( \approx 2278.69 \, \text{J} \)