A piece of luggage is being loaded onto an airplane by way of an inclined conveyor belt. The bag, which has a mass of 15.0 kg, travels 4.50 m up the conveyor belt at a constant speed without slipping. If the conveyor belt is inclined at a 30.0° angle, calculate the work done on the bag by: the force of gravity (Wg) the normal force (WN) the friction force (Wf) the conveyor belt (Wconveyor) the net force (Wnet)
1) Find the height the bag went up: 4.5sin30. Work is mgh
2)Normal force can NEVER do work
3)The friction force is not moving the bag
4)Work done by belt must be PE gained. So mgh again.
5)Net force is zero. Constant speed.
To calculate the work done on the bag by each force, we need to use the following formula:
Work (W) = Force (F) × Distance (d) × cosθ
where F is the magnitude of the force, d is the distance over which the force is applied, and θ is the angle between the force and the direction of motion.
1. Work done by the force of gravity (Wg):
The force of gravity acts vertically downwards and is given by the weight of the bag, which is the mass (m) multiplied by the acceleration due to gravity (g ≈ 9.8 m/s²):
Fg = m × g
Wg = Fg × d × cosθ
= (m × g) × d × cosθ
= (15.0 kg × 9.8 m/s²) × 4.50 m × cos(0°) [θ = 0° for the force of gravity as it acts vertically downwards]
= 661.5 J
Therefore, the work done on the bag by the force of gravity is 661.5 J.
2. Work done by the normal force (WN):
The normal force acts perpendicular to the inclined plane. Since the bag is not accelerating vertically, the normal force balances the gravitational force:
WN = Fg
Wn = Fg × d × cosθ
= (m × g) × d × cosθ
= (15.0 kg × 9.8 m/s²) × 4.50 m × cos(30°) [θ = 30° for the normal force as it is inclined at an angle of 30°]
= 661.5 J
Therefore, the work done on the bag by the normal force is also 661.5 J.
3. Work done by the friction force (Wf):
The friction force acts opposite to the bag's motion and is the force responsible for keeping the bag moving at constant speed. To calculate this force, we need to determine the magnitude of the force opposing motion, which is the product of the bag's mass, the acceleration due to gravity, and the coefficient of friction (μ):
Ff = m × g × μ
Wf = Ff × d × cosθ
= (m × g × μ) × d × cosθ
= (15.0 kg × 9.8 m/s² × μ) × 4.50 m × cos(30°) [θ = 30° for the friction force as it is inclined at an angle of 30°]
= 661.5 μ J
The coefficient of friction needs to be known in order to calculate the exact numerical value of the work done by the friction force (Wf). Please provide the coefficient of friction to determine the value.
4. Work done by the conveyor belt (Wconveyor):
Since the bag is moving up the conveyor belt at a constant speed without slipping, the work done by the conveyor belt is zero. This is because the force exerted by the conveyor belt is perpendicular to the direction of motion, so the angle θ between them is 90°, and cos 90° equals zero.
Wconveyor = 0 J
Therefore, the work done by the conveyor belt is zero.
5. Net work done on the bag (Wnet):
The net work done on the bag is the sum of the work done by all the forces acting on the bag:
Wnet = Wg + WN + Wf + Wconveyor
= 661.5 J + 661.5 J + Wf + 0 J
The value of Wf needs to be specified to calculate the net work done on the bag. Please provide the coefficient of friction to determine the net work done.