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
- Work done by the force \( P \): \( \approx 2889.12 , \text{J} \)
- Work done by the friction force: \( \approx -469.2 , \text{J} \)
- Work done by gravity: \( \approx -141.23 , \text{J} \)
- Total work done on the block: \( \approx 2278.69 , \text{J} \)