A rope with tension T = 150 N pulls a 15-kg block 3 m up an incline (alpha = 30 degrees). The coefficient of kinetic friction is k = 0,2. Find the work done by each force acting on the block.
To find the work done by each force acting on the block, we need to consider the forces involved and calculate the work done by each force separately. The forces acting on the block are:
1. The force of tension in the rope (T).
2. The force of gravity pulling the block downwards (mg).
3. The force of friction acting against the motion of the block up the incline (f).
Let's start with calculating the work done by the force of tension in the rope (T). The work done by a force is given by the formula:
Work = Force × Displacement × Cosine(angle between the force and displacement)
For the force of tension, the displacement is the distance the block moves up the incline, which is 3 meters. The angle between the force of tension and the displacement is 30 degrees, which is the angle of the incline.
So, the work done by the force of tension will be:
Work_Tension = T × 3 × cos(30°)
Work_Tension = 150 N × 3 m × cos(30°)
Work_Tension = 389.711 J (rounded to three decimal places)
Next, let's calculate the work done by the force of gravity (mg). The displacement in this case is also the vertical distance the block moves up the incline, which is 3 meters. The angle between the force of gravity and the displacement is 90 degrees, as gravity acts vertically downwards.
So, the work done by the force of gravity will be:
Work_Gravity = mg × 3 × cos(90°)
Work_Gravity = 15 kg × 9.8 m/s² × 3 m × cos(90°)
Work_Gravity = 0 J (since cos(90°) = 0)
Finally, let's calculate the work done by the force of friction (f). The work done by friction is given by the formula:
Work = -Frictional Force × Displacement
The frictional force is calculated using the coefficient of kinetic friction (k) and the normal force (N), which is the component of the force of gravity perpendicular to the incline.
Frictional Force (f) = k × N
The normal force (N) can be calculated using the formula:
N = mg × cos(θ)
where θ is the angle of the incline (30°).
So, N = 15 kg × 9.8 m/s² × cos(30°)
N ≈ 127.05 N (rounded to two decimal places)
Now, calculate the frictional force (f):
f = k × N
f = 0.2 × 127.05 N
f ≈ 25.41 N (rounded to two decimal places)
The displacement for the force of friction is also 3 meters since it acts parallel to the incline.
So, the work done by the force of friction will be:
Work_Friction = -f × 3
Work_Friction = -25.41 N × 3 m
Work_Friction = -76.23 J (negative because the force of friction is acting in the opposite direction of displacement)
In summary, the work done by each force acting on the block is:
- Work_Tension = 389.711 J
- Work_Gravity = 0 J
- Work_Friction = -76.23 J