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?

Answer 1

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**