A train: Power and friction

[envscigrl]

PROBLEM:
A train with a total mass 2.20E+6kg rises 600m as it travels a distance of 53.0km along a steady slope at a constant speed of 10.0km/hr. The frictional force on the train is 1.400 percent of the weight. Find the kinetic energy of the train.
I did this using K=.5*m*vsquared and got 8501240.
It then asks:

Find the energy dissipated by kinetic friction.

and

Find the power output of the train engines.

I am unsure as to how to find the energy dissipated by kinetic friction. I thought it involved using fk = muk * N
where Mu is the coefficient of friction and N is the normal force. Yet, I could not seem to find the answer.

For the power I tried using P= .5 m Vsquared / t. And finding t from the information given but this also did not work.

Thanks for helping!

 

[Doc Al]

The energy dissipated by friction is just the work done against friction. Note: They tell you what the frictional force is. No need to calculate it from mu and N.

The engine must provide power to provide the energy needed to raise the train against friction. Figure out how much energy is needed and how much time it takes. Ask yourself: Does the KE of the train change? Does the gravitational PE change? How much work is done against friction?

 

[wais]

To find the energy dissipated by kinetic friction, you can use the equation E=Fd, where E is the energy dissipated, F is the frictional force, and d is the distance traveled. In this case, the frictional force is 1.400 percent of the weight, so F=0.014*mg, where m is the mass of the train and g is the acceleration due to gravity. The distance traveled is 53.0km, so d=53000m. Plugging in these values, we get E=(0.014*2200000*9.8)*53000= 1.671E+9 J. This is the energy dissipated by kinetic friction.

To find the power output of the train engines, we can use the equation P=W/t, where P is power, W is work, and t is time. In this case, the work done by the engines is equal to the change in kinetic energy of the train. We already know the initial kinetic energy of the train (0 J), and we can calculate the final kinetic energy using the equation K=0.5mv^2. Plugging in the values, we get K=0.5*2200000*(10/3.6)^2= 2.420E+8 J. So, the work done by the engines is W=2.420E+8 J. The train travels at a constant speed of 10.0km/hr, which is equivalent to 2.78 m/s. Therefore, the time taken to travel the distance of 53.0km is t=53000/2.78= 19061 s. Plugging these values into the equation for power, we get P= (2.420E+8)/(19061)= 1.270E+4 W. This is the power output of the train engines.