Work of pulling sleigh in the top of the hill including the friction

In summary, to pull a loaded sleigh (total weight of 10 kg) up a 10 m high slope with a gradient of 30 degrees, the work required will depend on the linearly changing coefficient of friction between the sleigh and the ground. The parallel component of gravitational force and the force of friction must be taken into account, and the formula for elementary work is dW=mg(sinα+cosαμ)dx. To solve this problem, an expression for the friction coefficient that varies with position on the slope must be determined.
  • #1
Xaime
1
0
What work must be done to pull a loaded sleigh (total weight of 10 kg ) on the top height H=10 m. Gradient of slope amounts 30 degrees. Coefficient of friction between the sleigh and the ground decreases linearly along the way from 0,5 at the food of the hill to 0,1 in the top.

The pararell compnent of gravitational force
Code:
F _{s}=mg \sin \alpha
and the force of friction
Code:
F _{T}=mg\cos\alpha \mu

My formula for elementary Work is
Code:
dW=mg(\sin\alpha+ \cos \alpha \mu) dx

Now I have problem with integrate this equation, that is with the linearly changing coefficient of friction.
 
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  • #2
Xaime said:
What work must be done to pull a loaded sleigh (total weight of 10 kg ) on the top height H=10 m. Gradient of slope amounts 30 degrees. Coefficient of friction between the sleigh and the ground decreases linearly along the way from 0,5 at the food of the hill to 0,1 in the top.

The pararell compnent of gravitational force
Code:
F _{s}=mg \sin \alpha
and the force of friction
Code:
F _{T}=mg\cos\alpha \mu

My formula for elementary Work is
Code:
dW=mg(\sin\alpha+ \cos \alpha \mu) dx

Now I have problem with integrate this equation, that is with the linearly changing coefficient of friction.

Hi Xaime, Welcome to Physics Forums.

In future, please retain and use the Homework Posting Template that is provided when you start a new thread in the Homework sections of the Forums.

For your problem, you'll have to write an expression for the friction coefficient that varies appropriately with the position on the slope. So μ(x) = some function of x, where x is the distance along the slope.
 

Related to Work of pulling sleigh in the top of the hill including the friction

1. What is the purpose of pulling a sleigh in the top of the hill?

The purpose of pulling a sleigh in the top of the hill is to overcome the force of gravity and reach the bottom of the hill. This is known as potential energy conversion to kinetic energy.

2. How does friction affect the work of pulling a sleigh in the top of the hill?

Friction is a resistive force that acts against the motion of the sleigh. It can make the work of pulling the sleigh more difficult as it requires more force to overcome the frictional force.

3. Is the work of pulling a sleigh in the top of the hill affected by the weight of the sleigh?

Yes, the weight of the sleigh plays a role in the amount of work required to pull it up the hill. The heavier the sleigh, the more force is needed to overcome the force of gravity and friction.

4. How does the angle of the hill affect the work of pulling a sleigh?

The steeper the hill, the more work is required to pull the sleigh to the top. This is because the force of gravity is acting at a steeper angle, making it more difficult to overcome.

5. What other factors can affect the work of pulling a sleigh in the top of the hill?

Other factors that can affect the work of pulling a sleigh in the top of the hill include the condition and type of surface the sleigh is being pulled on. A rough or uneven surface can increase the amount of friction and make the work more difficult. Additionally, the strength and energy of the person or animal pulling the sleigh can also impact the amount of work required.

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