Calculating the Efficiency and Speed of a Toy Train

[ProPM]

A 0.50 kg battery-operated toy train moves at a constant velocity of 0.30 m/s along a level track. The power developed by the motor is 2.0 W and the total force opposing the motion of the train is 5.0 N

a) Calculate the efficiency of the train's motor
b) Assuming the efficiency and the opposing forces remain the same calculate the speed of the train as it climbs up an incline of 10 degrees to the horizontal

My answers:
a) P = F x v, 5 x 0.30, P = 1.5
(1.5 / 2) x 100 = 75%

b) Opposing force: W sin 10 + 5, = 5.9 N

0.75 = 5.9 x v / 2

v = 0.25

My question is: My teacher told me there were two ways about the second part of this problem. One of them is the one I did, but I forgot how to do the other one! I remember it was using GPE: So that, the Power = m x g x h, the h I think, being the velocity up the ramp x sin of 10...

Anyone can help me?

Thanks,
Pro PM

 

[anathema]

Hi Pro PM,

Thank you for sharing your answers and question with us. I can definitely help you with the second part of the problem using the formula for gravitational potential energy (GPE). Let's break it down step by step.

First, let's determine the GPE of the train at the bottom of the incline. We can use the formula GPE = mgh, where m is the mass of the train (0.50 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height of the incline. Since the train is on a level track, the height of the incline is 0. Therefore, the GPE at the bottom is 0.

Next, we need to calculate the GPE at the top of the incline. This is where the train reaches its maximum height, so h is equal to the length of the incline. To find this, we can use the formula h = l sinθ, where l is the length of the incline and θ is the angle of the incline (10 degrees). Plugging in the values, we get h = 0.087 m.

Now, we can use the formula for GPE again to find the GPE at the top of the incline. GPE = mgh = 0.50 x 9.8 x 0.087 = 0.426 J. This is the amount of energy that the train has gained by climbing up the incline.

Finally, we can use the formula for efficiency (efficiency = work output / work input) to find the speed of the train at the top of the incline. The work output is the GPE gained by the train (0.426 J) and the work input is the power developed by the motor (2.0 W). Plugging in the values, we get:

efficiency = 0.426 / 2.0 = 0.213

Now, we can use this efficiency to find the speed of the train at the top of the incline. We know that the power developed by the motor (2.0 W) is equal to the work done by the motor (force x distance) in a given time (1 second). So, we can rearrange the formula to find the force required to move the train up the incline:

Force = Power / distance = 2.0 / 0