A piece of luggage is being loaded onto an airplane by way of an inclined conveyor belt. The bag, which has a mass of 20.0 kg, travels 6.50 m up the conveyor belt at a constant speed without slipping. If the conveyor belt is inclined at a 40.0° angle, calculate the work done on the bag by: The force of gravity (g), the normal force (N), the friction force (f), the conveyor belt, and the net force.
Fg = m*g = 20kg * 9.8N/kg = 196 N.
Fp = 196*sin40 = 126 N. = Force parallel to incline.
Fn = 196*cos40 = 150 N. = Normal.
h = L*sin40 = 6.5*sin40 = 4.18 m.
Wg = Fg * h = 196 * 4.18 = 819.3 Joules.
Wn = = Fn * h = 150 * 4.18 = 628 N.
Fp-Ff = m*a = m*0 = 0
126-Ff = 0
Ff = 126 N.= Force of friction.
Wf = Ff * L = 126 * 6.5 = 819 Joules.
To calculate the work done on the bag by different forces, we need to understand the different forces acting on it.
1. Force of Gravity (G):
The force of gravity acting on the bag is given by the equation G = mg, where m is the mass of the bag and g is the acceleration due to gravity (approximately 9.8 m/s²). Therefore, the force of gravity is G = (20.0 kg) * (9.8 m/s²).
2. Normal Force (N):
The normal force is the force exerted by a surface to support the weight of an object resting on it. When the bag is on the inclined conveyor belt, the normal force acts perpendicular to the conveyor belt (opposite to the gravitational force). The magnitude of the normal force is N = mg*cos(θ), where θ is the angle of inclination (40.0°). Therefore, the normal force is N = (20.0 kg) * (9.8 m/s²) * cos(40.0°).
3. Friction Force (F):
The friction force acts parallel to the surface and opposes the motion of the bag. As the bag moves at a constant speed without slipping, the frictional force exactly balances the force that would cause slipping. The magnitude of the friction force is given by F = μ*N, where μ is the coefficient of friction between the bag and the conveyor belt. However, as the problem does not provide the coefficient of friction, we cannot calculate the exact value of the friction force.
4. Conveyor Belt:
The conveyor belt provides the force required to move the bag up the incline. This force is directed parallel to the incline and is equal to the friction force (assuming no other external forces are acting on the bag).
5. Net Force:
The net force acting on the bag is the vector sum of all the forces acting on it. In this case, the net force is equal to zero since the bag is moving at a constant speed without any acceleration. This means that the sum of the forces in the x-axis and y-axis will be zero.
Note: Without the coefficient of friction or any other information, we cannot calculate the exact work done by the friction force or the conveyor belt. We can only discuss their relationship in terms of magnitude.