To solve the problem step by step, we will analyze the forces acting on the block and calculate the required work done by different forces.
### Given Data:
- Mass of the block, \( m = 10 \, \text{kg} \)
- Length of incline, \( L = 5 \, \text{m} \)
- Height of incline, \( h = 3 \, \text{m} \)
- Coefficient of friction, \( \mu = 0.1 \)
- Gravitational acceleration, \( g = 10 \, \text{m/s}^2 \)
### Step 1: Calculate the angle of the incline
Using the right triangle formed by the height and length of the incline:
\[
\sin \theta = \frac{h}{L} = \frac{3}{5} \quad \text{and} \quad \cos \theta = \frac{4}{5}
\]
### Step 2: Calculate the gravitational force components
The gravitational force acting on the block is:
\[
F_g = mg = 10 \times 10 = 100 \, \text{N}
\]
The components of the gravitational force along the incline:
\[
F_{g \parallel} = mg \sin \theta = 100 \times \frac{3}{5} = 60 \, \text{N}
\]
\[
F_{g \perpendicular} = mg \cos \theta = 100 \times \frac{4}{5} = 80 \, \text{N}
\]
### Step 3: Calculate the normal force
The normal force \( N \) is equal to the perpendicular component of the gravitational force:
\[
N = F_{g \perpendicular} = 80 \, \text{N}
\]
### Step 4: Calculate the frictional force
The frictional force \( F_f \) opposing the motion is given by:
\[
F_f = \mu N = 0.1 \times 80 = 8 \, \text{N}
\]
### Step 5: Calculate the force exerted by the man
Since the block is sliding down at constant speed, the net force along the incline must be zero:
\[
P + F_f = F_{g \parallel}
\]
Thus, the force exerted by the man \( P \) is:
\[
P = F_{g \parallel} - F_f = 60 - 8 = 52 \, \text{N}
\]
### Step 6: Calculate the work done by the man
The work done by the man \( W_m \) is:
\[
W_m = P \cdot d \cdot \cos(180^\circ) = 52 \cdot 5 \cdot (-1) = -260 \, \text{J}
\]
### Step 7: Calculate the work done by gravity
The work done by gravity \( W_g \) is:
\[
W_g = F_{g \parallel} \cdot d = 60 \cdot 5 = 300 \, \text{J}
\]
### Step 8: Calculate the work done by the surface (friction)
The work done by friction \( W_f \) is:
\[
W_f = F_f \cdot d \cdot \cos(180^\circ) = 8 \cdot 5 \cdot (-1) = -40 \, \text{J}
\]
### Step 9: Calculate the work done by the resultant forces
Since the block moves with constant velocity, the net work done by the resultant forces \( W_r \) is:
\[
W_r = 0 \, \text{J}
\]
### Step 10: Calculate the change in kinetic energy
Since the velocity is constant, the change in kinetic energy \( \Delta KE \) is:
\[
\Delta KE = 0 \, \text{J}
\]
### Summary of Results:
(a) Work done by the man on the block: \( -260 \, \text{J} \)
(b) Work done by gravity on the block: \( 300 \, \text{J} \)
(c) Work done by the surface on the block: \( -40 \, \text{J} \)
(d) Work done by the resultant forces on the block: \( 0 \, \text{J} \)
(e) Change in K.E. of the block: \( 0 \, \text{J} \)
