Efficiency of a Cart Rolling Down an Inclined Plane

[xRadio]

Homework Statement

Alright for this thing, we had to pull a cart up a inclined plane and time how long it took to get to the bottom.

Distance from ground to top of inclined plane is - 0.125m
Length of Board is - 1.21m
Mass of cart - 1.09kg
Force to pull cart up - 1.4N
Time it took for cart to roll down board - 1.92s

So I need to calculate:
Energy Input (pulling cart up)
Energy Output
Efficiency

Homework Equations

W = Fd

Eff = Useful output energy/Input Energy x 100%

The Attempt at a Solution

So I am guessing Energy Input is
W=FD
W=(1.4)(1.21)
W = 1.694 J

Would Energy output be
E = 1/2mv^2?

 

[PhanthomJay]

I don't think so, wouldn't that be counting your losses twice? It takes 1.7 joules of input energy (Work) to pull the cart up to the top of the plane. How much energy does it have when it gets there?

 

[m2003]

First, it is important to note that the efficiency of a system is defined as the ratio of useful output energy to input energy, multiplied by 100%.

To calculate the efficiency of the cart rolling down the inclined plane, we need to first determine the useful output energy and the input energy.

The useful output energy in this case is the potential energy that is converted into kinetic energy as the cart rolls down the incline. This can be calculated using the formula E = mgh, where m is the mass of the cart, g is the acceleration due to gravity (9.8 m/s^2), and h is the height of the incline.

E = (1.09 kg)(9.8 m/s^2)(0.125 m) = 1.35 J

The input energy is the work done to pull the cart up the incline. This can be calculated using the formula W = Fd, where F is the force applied to the cart and d is the distance the cart is pulled up the incline.

W = (1.4 N)(1.21 m) = 1.694 J

Now, we can calculate the efficiency using the formula:

Eff = (Useful output energy/Input energy) x 100%

Eff = (1.35 J/1.694 J) x 100% = 79.7%

Therefore, the efficiency of the cart rolling down the inclined plane is approximately 80%. This means that 80% of the input energy (pulling the cart up the incline) was converted into useful output energy (the cart's kinetic energy as it rolled down the incline). The remaining 20% of the input energy was likely lost due to friction and other factors.