Potential & Kinetic Energy

... is lifting a mass of 200kg, up to a height of 2 metres. We have already seen how to calculate the potential energy of his weights:

Potential energy = work done = weight x height lifted

But here on Earth, weight (in N) = mass x 10 so:

Gravitational P.E = Mass g height (joules) (kg) (N/kg) (m)

(g has a different value on other planets)"

The book also tells me the formula for kinetic energy is:

K.E = ½ x mass x velocity squared K.E = ½mv2

Knowing this we can write:

P.E = K.E mgh = ½mv2 The formula can be simplified 20h = v2 SQRT(20h) = v

This formula will give us the average velocity for the trolley going down a ramp of h metres high. Once we have found this we can actually use the equation for average speed to find out how long it will take the trolley to reach the finish line and actually produce a theoretical result prior to conducting the experiment. Obviously, this won´t be necessary for a simple prediction, but it shows that the higher the ramp is raised, the higher the velocity of the trolley will be resulting in a quicker time to reach the finish line. I can also predict from this formuIa, the shape of the graph v against h. As h increases uniformly, by lets say 10cm each time, v will increase too - but not in proportion. This is due to the square root in the formula that we have to use to find v. The higher the height goes, the less gap there will be between the velocity of the present and previous heights. The graph will look something like this:

Therefore, I predict

Increase in height of ramp = Increase in velocity of trolley

Secondary Experiment Again, for the secondary experiment, we just need to examine the equation that states potential energy at he top equals the kinetic energy at the bottom.

P.E = K.E Mgh = K.E Now looking at the equations at this stage, it seems sensible to say that a larger mass will result in more kinetic energy, and hence a faster velocity. But lets look at the formula for kinetic energy. Mgh = ½mv2 Now we can see here that although a larger mass will indeed result in a larger amount of potential, and therefore kinetic, energy it will not result in higher velocity. BOTH sides of the equation contain mass, which simply means they cancel each other out. Gh = ½v2

Therefore I predict that there will be no significant change in velocity when the weight of the trolley is altered.

SKILL AREA O : OBTAINING EVIDENCE

This section is mainly putting our planning into action, and hence is nearly all practical work so not much written work will be produced.

Primary Experiment When we came to conduct our experiment, we decided to alter our plan and do two experiments. One using a stop-watch timer and one using a light gate to record the velocity of the trolley for more accuracy.

Manually timing the experiment: Height of runway (cm) Time taken to travel 2m (sec) Velocity [distance/time] (m/s) Average speed (m/s) 10cm 3.42 3.58 3.39 0.58 0.56 0.59 0.58 20cm 2.23 2.15 2.09 0.9 0.93 0.9 0.91 30cm 1.81 1.75 1.64 1.11 1.14 1.22 1.17 40cm 1.39 1.52 1.37 1.43 1.32 1.46 1.41 50cm 1.24 1.25 1.28 1.61 1.6 1.56 1.59

Using a light gate and computer software: Height of runway (cm) Speed (m/s) Average speed (m/s) 10cm ...