Solving Highway Hill Work Problem: Find v_down

[NasuSama]

Homework Statement

A highway goes up a hill, rising at a constant rate of 1.00 m for every 40 m along the road. A truck climbs this hill at constant speed v_up = 18 m/s, against a resisting force (friction) f equal to 1/20 of the weight of the truck. Now the truck comes down the same hill, using the same power as it did going up. Find v_down, the constant speed with which the truck comes down the hill.

ASSUME: the resisting force (friction) has the same magnitude going up as going down.

Homework Equations

Hm...

→W = mgd
→KE = W = ½ * mv²

The Attempt at a Solution

Tried to use this form:

½ * m * v_up² - ½ * m * v_down²

BUT doesn't seem to find the correct path.

 

[cheahchungyin]

Didn't calculate in the form of work done, resolved the forces.

First, The equation when the truck climbs the hill. (take power=P)

P/v_up=(1/20)m+mg sin(tan^-1(1/40))

Then, the equation when the truck is going down.

P/v_down=(1/20)m-mg sin(tan^-1(1/40))

divide the first equation by the second equation to eliminate the unknowns.
You'll get the answer :)

Tell me if my equations are unclear

 

[NasuSama]

Didn't calculate in the form of work done, resolved the forces.

First, The equation when the truck climbs the hill. (take power=P)

P/v_up=(1/20)m+mg sin(tan^-1(1/40))

Then, the equation when the truck is going down.

P/v_down=(1/20)m-mg sin(tan^-1(1/40))

divide the first equation by the second equation to eliminate the unknowns.
You'll get the answer :)

Tell me if my equations are unclear

The equations seem to be incorrect. I enter in the solution, but not right.

 

[NasuSama]

First, The equation when the truck climbs the hill. (take power=P)

P/v_up=(1/20)m+mg sin(tan^-1(1/40))

P/v_down=(1/20)m-mg sin(tan^-1(1/40))

Tell me if my equations are unclear

How did you get mgsin(arctan(1/40)) by the way? I see that you use the kinetic formula, but it's not right.

 

[cheahchungyin]

eh? should be correct, I can't see where I did wrong >.<
P/v_up is the constant force applied by the truck.
When it is going up, (1/20)m is the friction and mg sin(tan-1(1/40)) is the resultant gravitational force acting on the truck.
When going down, direction of resultant gravitational force is the same as the constant force. So I put a negative sign there.
I can't see what I did wrong.

 

[cheahchungyin]

Sory I just noticed my careless mistake :P .
The resistant should be (1/20)mg instead of (1/20)m.
try it, should be correct this time.

 

[NasuSama]

Sory I just noticed my careless mistake :P .
The resistant should be (1/20)mg instead of (1/20)m.
try it, should be correct this time.

What about v²? Does that take in account?

 

[cheahchungyin]

v2? you mean kinetic energy right? I'm resolving everything in forces, not energy form, so it doesn't take into account