A 975 N object is pulled at a constant velocity up an incline with a force of 327 n parallel to the incline as shown in the diagram. Assume that the difference between the work output and the work input is the work required to overcome friction. a) what is the force of friction along the incline? the length of the incline is 7 m and the height of the incline is 2.22 m. b) what is the efficiency of the system?
To find the force of friction along the incline, we can start by calculating the work input and work output.
a) Work input is the force applied parallel to the incline (327 N) multiplied by the distance along the incline (7 m). Therefore, the work input is:
Work input = Force × Distance = 327 N × 7 m = 2289 N·m
Since the object is pulled at a constant velocity, the work output is equal to the work input. Thus, the work output is also 2289 N·m.
The work output is the force exerted on the object parallel to the incline (which helps it move up the incline) multiplied by the distance along the incline (7 m). Since the object is moving at a constant velocity, the net force acting on it must be zero. Therefore, the force parallel to the incline is equal in magnitude and opposite in direction to the force of friction.
The force of friction can be calculated using the formula:
Force of friction = Work output / Distance
Force of friction = 2289 N·m / 7 m = 327 N
Therefore, the force of friction along the incline is 327 N.
b) The efficiency of the system can be calculated using the formula:
Efficiency = (Useful output energy / Input energy) × 100%
In this case, the useful output energy is the work output, which is 2289 N·m, and the input energy is the work input, which is also 2289 N·m. Substituting these values into the formula, we get:
Efficiency = (2289 N·m / 2289 N·m) × 100% = 100%
Therefore, the efficiency of the system is 100%.
a) To find the force of friction along the incline, we need to use the work-energy principle. The work done by the force of gravity is equal to the work done by the applied force parallel to the incline and the work done against friction.
The work done by the force of gravity is given by:
Work_gravity = force_gravity x distance x cos(angle)
The work done by the applied force parallel to the incline is given by:
Work_applied = force_applied x distance x cos(angle)
Since the object is pulled at a constant velocity, the net work done must be zero. Therefore, the work done against friction is:
Work_friction = - (Work_gravity + Work_applied)
Let's calculate each term step-by-step:
1. Calculate the force of gravity:
force_gravity = mass x gravity
Given: mass = 975 N and gravity = 9.8 m/s^2
force_gravity = 975 x 9.8 N
2. Calculate the angle with respect to the horizontal:
tan(angle) = height / length
Given: height = 2.22 m and length = 7 m
angle = arctan(2.22 / 7)
3. Calculate Work_gravity:
Work_gravity = force_gravity x distance x cos(angle)
4. Calculate Work_applied:
Work_applied = force_applied x distance x cos(angle)
Given: force_applied = 327 N
5. Calculate Work_friction:
Work_friction = - (Work_gravity + Work_applied)
Finally, the force of friction along the incline can be calculated by dividing Work_friction by the distance.
b) The efficiency of the system can be calculated by dividing the work output by the work input. We can determine the work output by multiplying the force applied by the distance, and the work input by multiplying the force of gravity by the same distance. The efficiency is given by:
efficiency = work_output / work_input
Let's calculate both parts one by one.