Work done down an inclined plane

In summary, the conversation discusses the calculation of work, power, and efficiency for a toy car moving down a slope under the influence of gravity and friction. The work done is the sum of the work done by both forces, and the net work done on the car will determine its efficiency. The problem may require the calculation of the work done by gravity specifically for efficiency purposes.
  • #1
Dongorgon
20
0

Homework Statement


A toy car of mass 20kg moves down a slope of length 35m, which is inclined at arcsin(2/7) to the horizontal. The height of the car initially is 10m, and a constant frictional force of 100N acts against the car through motion. Calculate the work done, power and efficiency of the system:


The Attempt at a Solution


My initial issue here is calculating the work done by the car traveling down the entirety of the slope. I'm not sure whether the work done is just the friction force x distance which would give 3500J, or whether you need to take into account the horizontal component of the cars weight when descending the slope, resolving horizontally?
Any advice here would be greatly appreciated! Thanks.
 
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  • #2
Dongorgon,

It doesn't make sense to speak of the work done "by" the car. Work is force * distance, so work is done BY forces ON objects as those forces move those objects over distances. If you want to compute the total work done ON the car, you have to take into account the work done by all forces acting on the car. So you have to account for the work done by gravity as well as the work done by friction. Another way to look at it: the total (net) work done on the car will be the work done by the net force. I hope this helps.
 
  • #3
You need to take into account both gravity and friction. The work done by friction is the force, 100N, times the distance traveled, 35m. The work done by gravity can be calculated by the decrease in potential energy, mg= (20)(9.81) times the vertical distance, (2/7)35= 10 m, the car has gone. Becareful- one is work done increasing the speed of the car, the other is work done decreasing the speed of the car.

(There is no "horizontal component of weight". Weight is a force vector pointing vertically.)
 
  • #4
cepheid said:
the total (net) work done on the car will be the work done by the net force. I hope this helps.

Given the question about efficiency, I suspect they want the work done by gravity.
Dongorgon, have you quoted the problem exactly as given?
 
  • #5


I would approach this problem by first breaking it down into its components and then using relevant equations to solve for the work done, power, and efficiency of the system.

Firstly, we can calculate the total distance traveled by the car using trigonometry. The horizontal distance traveled by the car is 35m x cos(arcsin(2/7)) = 31.6m. This is the distance over which the frictional force of 100N is acting. We can then use the formula W = Fd to calculate the work done by the frictional force, which is equal to 100N x 31.6m = 3160J.

Next, we need to consider the work done by the car's weight as it descends the slope. This can be calculated using the formula W = mgh, where m is the mass of the car (20kg), g is the acceleration due to gravity (9.8m/s^2), and h is the vertical distance traveled (10m). Therefore, the work done by the car's weight is 20kg x 9.8m/s^2 x 10m = 1960J.

The total work done by the system is then the sum of these two values, which is 3160J + 1960J = 5120J.

To calculate the power of the system, we use the formula P = W/t, where W is the work done (5120J) and t is the time taken for the car to travel down the slope. Unfortunately, the time is not given in the problem, so we cannot calculate the power.

Lastly, to calculate the efficiency of the system, we use the formula efficiency = (useful output energy / total input energy) x 100%. In this case, the useful output energy is the work done by the car's weight (1960J) and the total input energy is the work done by the system (5120J). Therefore, the efficiency is (1960J / 5120J) x 100% = 38.3%.

In conclusion, the work done by the system is 5120J, the power cannot be calculated without the time taken, and the efficiency is 38.3%. It is important to consider all the components of the system and use relevant equations to calculate the desired values. I hope this helps!
 

1. What is work done down an inclined plane?

Work done down an inclined plane refers to the amount of force required to move an object from one point to another on a slanted surface. It takes into account both the distance the object is moved and the force needed to move it against the force of gravity.

2. How is work calculated down an inclined plane?

To calculate work done down an inclined plane, you can use the formula W = Fdcosθ, where W is the work, F is the force applied, d is the distance moved, and θ is the angle of the incline.

3. Why is less work required to move an object down an inclined plane compared to lifting it vertically?

This is because the force required to move an object down an inclined plane is less than the force required to lift it vertically. This is due to the component of the force acting against the force of gravity being reduced when moving down an incline.

4. What is the relationship between the angle of the incline and the work done down an inclined plane?

The relationship between the angle of the incline and the work done is directly proportional. This means that as the angle of the incline increases, so does the work required to move the object down the incline.

5. How does friction affect the work done down an inclined plane?

Friction is a resistive force that acts against the movement of an object. It increases the amount of work required to move an object down an inclined plane as it opposes the force applied. Therefore, the higher the coefficient of friction, the more work is needed to overcome it.

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