Calculating constant velocity using work

[anonymous12]

Homework Statement

A 25.6-kg boy pulls a 4.81-kg toboggan up a hill inclined at 25.7º to the horizontal. The vertical height of the hill is 27.3 m and the coefficient of friction of the hill surface is 0.4
a) Determine how much work the boy must apply on the toboggan to pull it at a constant velocity up the hill? (Hint: An FBD is needed to see all your forces acting on te toboggan)

Homework Equations

W = F x ∆d
Eg = mgh
Ek = 0.5mv^2
vf^2 - vi^2 = 2a∆d

The Attempt at a Solution

-mgsin25.7º - \mu Kmgcos25.7º = m * a
-(30.41 x 9.8 x sin 25.7º) - (.4 x 30.41 x 9.8 x cos 25.7) = 30.41 * a
-129.24 - 106.46 = 30.41a
a = -7.75 m/s^2Calculating length of the hill:

\frac{27.3}{sin25.7º} = 62.95m

I don't really know what to do after this.

 

[cepheid]

Homework Statement

A 25.6-kg boy pulls a 4.81-kg toboggan up a hill inclined at 25.7º to the horizontal. The vertical height of the hill is 27.3 m and the coefficient of friction of the hill surface is 0.4
a) Determine how much work the boy must apply on the toboggan to pull it at a constant velocity up the hill? (Hint: An FBD is needed to see all your forces acting on te toboggan)

Homework Equations

W = F x ∆d
Eg = mgh
Ek = 0.5mv^2
vf^2 - vi^2 = 2a∆d

The Attempt at a Solution

-mgsin25.7º - \mu Kmgcos25.7º = m * a
-(30.41 x 9.8 x sin 25.7º) - (.4 x 30.41 x 9.8 x cos 25.7) = 30.41 * a
-129.24 - 106.46 = 30.41a
a = -7.75 m/s^2Calculating length of the hill:

\frac{27.3}{sin25.7º} = 62.95m

I don't really know what to do after this.

If the boy is pulling the toboggan up the hill at a constant speed, then what must the acceleration be equal to? Hint: your mistake was that you forgot one of the forces in the force balance equation: the applied force from the boy!

 

[LabGuy330]

The reply above is spot on.

reconsider your acceleration and add the pulling force from the boy.

 

[anonymous12]

Ooooh thank you! I feel like an idiot now :(: