Using Energy to Solve Work Problem

[murrskeez]

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

f_k = E_f - E_i
f_k = μ_k*F_N
F_N=mgcos15°

The Attempt at a Solution

First, I found the work done by the friction force: W_fk = S(μ_kF_N)cos180°
W_fk = μ_k(mgcos15°)(-1)

Then I solved for the work done by friction using the energy equation:
W = 1/2(mv²) +mgy - 1/2(mv_i²) -mgy_i
W = 0 + mgy - 1/2(mv_i²) - 0

then I set the two equations equal to one another:
μk(mgcos15°)(-1) = mgy - 1/2(mv_i²)

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

y = μk(gcos15°)(-1) + 1/2(v_i²) / 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.

 

[PeterO]

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

f_k = E_f - E_i
f_k = μ_k*F_N
F_N=mgcos15°

The Attempt at a Solution

First, I found the work done by the friction force: W_fk = S(μ_kF_N)cos180°
W_fk = μ_k(mgcos15°)(-1)

Then I solved for the work done by friction using the energy equation:
W = 1/2(mv²) +mgy - 1/2(mv_i²) -mgy_i
W = 0 + mgy - 1/2(mv_i²) - 0

then I set the two equations equal to one another:
μk(mgcos15°)(-1) = mgy - 1/2(mv_i²)

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

y = μk(gcos15°)(-1) + 1/2(v_i²) / 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° ?

 

[murrskeez]

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?

 

[frogjg2003]

Where did s go in your W_fk 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.

 

[murrskeez]

Where did s go in your W_fk 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 W_other = 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?

 

[frogjg2003]

Go back a step or two and write out the W_fk=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.

 

[murrskeez]

Go back a step or two and write out the W_fk=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

 

[frogjg2003]

You're welcome.