How Steep is the Hill if a Car Uses More Power Climbing It?

[curly_ebhc]

Homework Statement

There is a car of mass 1900kg that travels at a steady speed of 27 m/s up and down a hill. The force of friction is the same in both directions and it takes 49 more horsepower to go up the hill than down the hill. What is the inclination of the hill.

Given: mass of car= 1900kg , velocity = 27 m/s, difference in power =49 hp =35000 W.

I set it up as an energy problem. First I got rid of time and set that for every second the car traveled 27 m and the difference in the (change in) energy was 35000 J.

Homework Equations

The Attempt at a Solution

Step 1: I set ∆PE(up) -∆PE(down) = 35000J
Then 2∆PE(up)=35000J

Step 2: I used potential energy

h= 35000J/[2(1900kg)(10 m/s2) = .92 m

Step 3: A set up a right triangle: hypotenuse=27m and height=.92m

I used theta=arcsine (.92m/27m) = 1.95 Degrees.

I don’t know the answer but I think that I did everything correctly. Can anyone find a mistake?

 

[Redbelly98]

Looks right to me!

You might want to give the final answer to 2 sig figs though.

Good job.

 

[thomate1]

Your approach to solving this energy problem is correct. However, there are a few minor errors in your calculations that may have resulted in an inaccurate answer.

In step 1, the equation should be ∆PE(up) - ∆PE(down) = 35000 J, as the difference in potential energy between going up and down the hill is equal to the amount of additional energy required to go up the hill. This means that the equation should be 2∆PE(up) = 35000 J, not 2∆PE(up) = 70000 J as you have written.

In step 2, your calculation for potential energy is correct, but the unit should be in meters, not Joules. So the calculation should be h = 35000 J / [2(1900 kg)(10 m/s^2)] = 0.92 m.

Step 3 is correct, but the calculated value for theta should be 1.95 degrees, not 0.52 degrees as you have written.

Overall, your approach is sound and your calculations are mostly correct, just a few minor errors in units and equations. Keep up the good work!