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:
- 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: \[ N = W \cos(37^\circ) = 121.5 \text{ N} \times \cos(37^\circ) \approx 121.5 \times 0.798 \approx 96.9 \text{ N} \]
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Calculate the frictional force: \[ F_{\text{friction}} = \mu N = 0.25 \times 96.9 \approx 24.2 \text{ N} \]
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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