To solve the problem step by step, we will analyze the forces acting on the block and calculate the required work done by each force.
### Given Data:
- Mass of the block, \( m = 10 \, \text{kg} \)
- Length of the incline, \( L = 5 \, \text{m} \)
- Height of the incline, \( h = 3 \, \text{m} \)
- Coefficient of friction, \( \mu = 0.1 \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)
### Step 1: Calculate the angle of incline
Using the triangle formed by the height and the length of the incline, we can find the sine and cosine of the angle \( \theta \):
\[
\sin \theta = \frac{h}{L} = \frac{3}{5}, \quad \cos \theta = \frac{\text{base}}{L} = \frac{4}{5}
\]
### Step 2: Calculate the normal force
The normal force \( N \) can be calculated using the vertical forces:
\[
N = mg \cos \theta = mg \left(\frac{4}{5}\right)
\]
Substituting the values:
\[
N = 10 \times 10 \times \frac{4}{5} = 80 \, \text{N}
\]
### Step 3: Calculate the frictional force
The frictional force \( f \) is given by:
\[
f = \mu N = 0.1 \times 80 = 8 \, \text{N}
\]
### Step 4: Calculate the gravitational force along the incline
The gravitational force acting down the incline is:
\[
F_g = mg \sin \theta = 10 \times 10 \times \frac{3}{5} = 60 \, \text{N}
\]
### Step 5: Calculate the force exerted by the man
Since the block is moving at constant speed, the net force along the incline must be zero:
\[
P + f - F_g = 0 \implies P = F_g - f = 60 - 8 = 52 \, \text{N}
\]
### Step 6: (a) Work done by the man on the block
The work done by the man is given by:
\[
W_m = P \cdot d \cdot \cos(0^\circ) = P \cdot d = 52 \cdot 5 = 260 \, \text{J}
\]
Since the force is acting against the displacement, the work done is negative:
\[
W_m = -260 \, \text{J}
\]
### Step 7: (b) Work done by gravity on the block
The work done by gravity is:
\[
W_g = F_g \cdot d \cdot \cos(180^\circ) = 60 \cdot 5 \cdot (-1) = -300 \, \text{J}
\]
### Step 8: (c) Work done by the surface on the block
The work done by the frictional force (surface) is:
\[
W_f = f \cdot d \cdot \cos(180^\circ) = 8 \cdot 5 \cdot (-1) = -40 \, \text{J}
\]
### Step 9: (d) Work done by the resultant forces on the block
Since the block is moving at constant speed, the net work done by the resultant forces is zero:
\[
W_{net} = 0 \, \text{J}
\]
### Step 10: (e) Change in K.E. of the block
Since the velocity is constant, the change in kinetic energy is:
\[
\Delta KE = KE_{final} - KE_{initial} = 0 \, \text{J}
\]
### Summary of Results:
- (a) Work done by the man: \( -260 \, \text{J} \)
- (b) Work done by gravity: \( -300 \, \text{J} \)
- (c) Work done by the surface: \( -40 \, \text{J} \)
- (d) Work done by resultant forces: \( 0 \, \text{J} \)
- (e) Change in K.E.: \( 0 \, \text{J} \)
