A 40.0 kg crate is pushed at a constant speed up a rough 6.0 m long ramp inclined at 30.0° with the horizontal. The force of friction between the crate and ramp is 6.0 N.Find the work done by the pushing force.
To find the work done by the pushing force, we need to determine the total distance over which the force is applied and the magnitude of the force itself.
1. Find the vertical distance the crate moves:
The given ramp is inclined at 30.0° with the horizontal, and the length of the ramp is 6.0 m.
The vertical distance the crate moves can be calculated using trigonometry:
Vertical distance = Length of the ramp * sin(30.0°)
= 6.0 m * sin(30.0°)
= 3.0 m
2. Find the force required to push the crate:
The force required to push the crate on an inclined surface can be determined by balancing the gravitational force and the force of friction. We have the following forces:
- Gravitational force (acting vertically downwards) = mass * acceleration due to gravity
- Force of friction (acting parallel to the inclined surface) = 6.0 N (given)
Since the crate is moving at a constant speed, the net force on the crate is zero. Therefore, the force required to push the crate up the ramp is counterbalanced by the gravitational force and the force of friction.
Gravitational force = mass * acceleration due to gravity
= 40.0 kg * 9.8 m/s²
= 392.0 N
Since the crate is moving at a constant speed, the force required to push the crate up the ramp is equal to the force of friction:
Force required = 6.0 N
3. Calculate the work done:
The work done by a force is given by the equation: Work = Force * Distance * cos(θ), where θ is the angle between the force and the direction of displacement.
In this case, the angle between the force and the displacement is 0° since the force and displacement are in the same direction. Therefore, cos(0°) = 1.
Work done = Force * Distance * cos(θ)
= 6.0 N * 6.0 m * cos(0°)
= 36.0 J
Therefore, the work done by the pushing force is 36.0 Joules.