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A confusing word problem. Can any one solve it?

  • #1

Homework Statement



A 240gram toy car is driven by an electric motor that has a constant power output. The car can climb a 24° incline at 16m/s and can travel on a horizontal table at 39m/s. The frictional force retarding the motion is kv, where k is a constant and v is its speed. How steep an incline can it climb with a speed of 28m/s?


The Attempt at a Solution



I've tried solving this for 2 days now and but not succeeded. It seems quite simple but it is the kv part that I don't know how to use.
Please could someone explain how to solve it?

Help is very much appreciated!
Thank you!
 

Answers and Replies

  • #2
berkeman
Mentor
57,497
7,524

Homework Statement



A 240gram toy car is driven by an electric motor that has a constant power output. The car can climb a 24° incline at 16m/s and can travel on a horizontal table at 39m/s. The frictional force retarding the motion is kv, where k is a constant and v is its speed. How steep an incline can it climb with a speed of 28m/s?


The Attempt at a Solution



I've tried solving this for 2 days now and but not succeeded. It seems quite simple but it is the kv part that I don't know how to use.
Please could someone explain how to solve it?

Help is very much appreciated!
Thank you!
Welcome to the PF.

The k*v part is just like wind resistance and/or rolling resistance. They are saying that the only retarding force on flat ground is this resistance, which is proportional to velocity.

Do the 3 FBDs for the 3 cases, and include this resistant force in each. You should end up with a set of equations that you can use to solve for the velocity in the 3rd situation. Please show us your work...
 
  • #3
Oh thank you :)
I still seem to be going in circles.

I made a table with the kinetic and potential energies for each one: (by the way the speeds are supposed to be in cm/s - i typed it wrong)

1) 0.39m/s --> Kinetic E = m(v^2)/2 = 0.018252 J
Potential E = mgh = 0 J

2) Kinetic = 0.003072 J
Potential = 0.1531 J

3) Kinetic = 0.009408 J
Potential = ?? (i don't know theta, that is what i need to find)

So i guess 2) is best to work with since i have both values so i can find friction:
so friction is PE - KE = 0.1531 - 0.003072 = 0.150028 J

I have no idea if this is right...
and how do i include the "motor has a constant power output" ?

0.39m/s
 
  • #4
PhanthomJay
Science Advisor
Homework Helper
Gold Member
7,156
502
so friction is PE - KE = 0.1531 - 0.003072 = 0.150028 J

I have no idea if this is right...
You are going about this wrong by trying to use energy concepts and incorrect equations. You need to draw free body diagrams as berkeman has suggested, and apply Newton's first law.
and how do i include the "motor has a constant power output" ?
Power is force times velocity, that is, P = Fv. The power delivered by the motor is the same for all cases. Write down all givens and relevant equations. Please show your work using free body diagrams.
 

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