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

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

[tex]\frac{27.3}{sin25.7º} = 62.95m[/tex]

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

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

[tex]\frac{27.3}{sin25.7º} = 62.95m[/tex]

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 :(:

 

[asteriatic]

To calculate the work done by the boy, we can use the formula W = F x ∆d, where W is the work, F is the force applied and ∆d is the displacement. In this case, the boy is pulling the toboggan at a constant velocity, so the net force on the toboggan must be zero. This means that the force the boy applies must be equal and opposite to the sum of the other forces acting on the toboggan, which are the force of gravity and the force of friction.

So, the work done by the boy can be calculated as:

W = (mgsin25.7º + \mu Kmgcos25.7º) * 62.95m

= (25.6 x 9.8 x sin 25.7º + 0.4 x 25.6 x 9.8 x cos 25.7) * 62.95

= 1657.15 J

Therefore, the boy must apply 1657.15 J of work to pull the toboggan at a constant velocity up the hill.