Using Energy to Solve Work Problem

AI Thread Summary
A truck traveling at 11.1 m/s down a 15° hill with locked brakes is analyzed to determine how far it travels before stopping, using energy principles. The work done by friction is calculated using the coefficient of kinetic friction (0.750) and the normal force. Initial calculations led to an incorrect distance of 26.7 m, while the correct answer is 13.5 m. Key discussions focused on correcting the angle used in the work equation and ensuring that all variables were properly represented in terms of distance. The final solution was reached after clarifying the relationship between distance and height in the energy equations.
murrskeez
Messages
13
Reaction score
0

Homework Statement



A truck is traveling at 11.1 m/s down a hill when the brakes on all four wheels lock. The hill makes an angle of 15.0° with respect to the horizontal. The coefficient of kinetic friction between the tires and the road is 0.750. How far does the truck travel before coming to a stop? Use energy to solve this problem.

Homework Equations



fk = Ef - Ei
fk = μk*FN
FN=mgcos15°

The Attempt at a Solution



First, I found the work done by the friction force: Wfk = S(μkFN)cos180°
Wfk = μk(mgcos15°)(-1)

Then I solved for the work done by friction using the energy equation:
W = 1/2(mv2) +mgy - 1/2(mvi2) -mgyi
W = 0 + mgy - 1/2(mvi2) - 0

then I set the two equations equal to one another:
μk(mgcos15°)(-1) = mgy - 1/2(mvi2)

the m's canceled out and I solved for y:

y = μk(gcos15°)(-1) + 1/2(vi2) / g

and plugged in values:

y = (0.750)(9.8m/s2)(cos15°)(-1) + 1/2(11.0)2 / 9.8m/s2

and got: y = 6.90 m

knowing that the truck was traveling at 15°, sin15° = y/s
s = 6.90/sin15°
s = 26.7m

the correct answer should be 13.5m. Any help/advice would be much appreciated.
 
Physics news on Phys.org
murrskeez said:

Homework Statement



A truck is traveling at 11.1 m/s down a hill when the brakes on all four wheels lock. The hill makes an angle of 15.0° with respect to the horizontal. The coefficient of kinetic friction between the tires and the road is 0.750. How far does the truck travel before coming to a stop? Use energy to solve this problem.



Homework Equations



fk = Ef - Ei
fk = μk*FN
FN=mgcos15°


The Attempt at a Solution



First, I found the work done by the friction force: Wfk = S(μkFN)cos180°
Wfk = μk(mgcos15°)(-1)

Then I solved for the work done by friction using the energy equation:
W = 1/2(mv2) +mgy - 1/2(mvi2) -mgyi
W = 0 + mgy - 1/2(mvi2) - 0

then I set the two equations equal to one another:
μk(mgcos15°)(-1) = mgy - 1/2(mvi2)

the m's canceled out and I solved for y:

y = μk(gcos15°)(-1) + 1/2(vi2) / g

and plugged in values:

y = (0.750)(9.8m/s2)(cos15°)(-1) + 1/2(11.0)2 / 9.8m/s2

and got: y = 6.90 m

knowing that the truck was traveling at 15°, sin15° = y/s
s = 6.90/sin15°
s = 26.7m

the correct answer should be 13.5m. Any help/advice would be much appreciated.

Should that 180° angle you used have actually been 15° ?
 
I thought that the in the formula W = sfcos∅ , the angle ∅ should represent the angle between the direction of s and the corresponding force, which in this case would be friction...and those two directions are 180° apart, right?
 
Where did s go in your Wfk equation? Also, what is the first equation in the relevant equations section? Don't use y in your potential energy equation, write in terms of s.
 
frogjg2003 said:
Where did s go in your Wfk equation? Also, what is the first equation in the relevant equations section? Don't use y in your potential energy equation, write in terms of s.

Oops, looks like I dropped the s. Looks like that would make

y = sμk(gcos15°)(-1) + 1/2(11.0)2 / 9.8 m/s2

could I then relate s and y through sin? Am I even on the right track?

The first equation is Wother = ΔE , other in this case meaning friction. Was given to me by my professor. But s and y are representing two different things, and y is a reflection of the incorporation of gravitational potential energy into the problem so I don't see how I can get around using it?
 
Go back a step or two and write out the Wfk=ΔKE+ΔPE equation. Don't use y, use s and θ. Also, keep a close eye on the signs. s is downwards, but both ΔEs should be negative.
 
frogjg2003 said:
Go back a step or two and write out the Wfk=ΔKE+ΔPE equation. Don't use y, use s and θ. Also, keep a close eye on the signs. s is downwards, but both ΔEs should be negative.

Finally got the correct answer, thank you very very much :smile:
 
You're welcome.
 
Back
Top