Gravitational and kinetic energy problem

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
10 replies · 3K views
lzh
Messages
109
Reaction score
0

Homework Statement


Given: 1 hp = 746 W and 1 m/s =2.237 mi/h.
When an automobile moves with constant
velocity, the power developed in a certain en-
gine is 85.4 hp.
What total frictional force acts on the car
at v = 72.8 mi/h? Answer in units of N.


Homework Equations


Kinetic Energy=.5mv^2
Disspated energy=F*displacement <-the f would be the frictional force.
W=change in energy(J)/change in time(s)

The Attempt at a Solution


i first converted 85.4 hp to W: 63708.4W
then the v to m/s: 32.54m/s

i would like to find the amount of joules in 63708.4W, but I don't know the time. So I probably need to find a way to cancel it.

also, I don't know the mass of the car...nor the displacement. I'm not sure of what to do with so many unknowns.
Can someone point my problems out?
Ty in advance!
 
Physics news on Phys.org
Why should you want joules for??

Instead, how are the quantities power, force and velocity related?
 
power=Force*displacement/time
after some simplification
I'm starting to see something though
edit:i forgot to type /t
 
Last edited:
You have constant velocity. What is therefore displacement/time equal to?
 
Last edited:
so that means Force*v=63708W
substituting v i get:
F=1957.84N
but that's not the frictional force, I'm going to solve for that next
 
So, what you have found is the force the engine provides on the system, agreed?

Now, how much acceleration does the car experience if it moves with constant velocity?
 
yeah that's the force of the engine.
car at constand velocity has 0 accelaeration
so at constant velocity, the force of friction must be equal to force of engine?
so F=1957.84N is correct?
 
Last edited:
Correct!:smile:
 
ty!
it was correct, according to my utexas's homework service
 
lzh said:
ty!
it was correct, according to my utexas's homework service

How fortunate for the homework service! :smile: