A box of mass m𝑚 is sliding along a horizontal surface.Part AThe box leaves position x=0𝑥=0 with speed v0𝑣0. The box is slowed by a constant frictional force until it comes to rest at position x=x1𝑥=𝑥1.Find Ff𝐹f, the magnitude of the average frictional force that acts on the box. (Since you don't know the coefficient of friction, don't include it in your answer.)Express the frictional force in terms of m𝑚, v0𝑣0, and x1𝑥1.View Available Hint(s)for Part AActivate to select the appropriates template from the following choices. Operate up and down arrow for selection and press enter to choose the input value typeActivate to select the appropriates symbol from the following choices. Operate up and down arrow for selection and press enter to choose the input value typeFf𝐹f =nothingSubmitPart BAfter the box comes to rest at position x1𝑥1, a person starts pushing the box, giving it a speed v1𝑣1.When the box reaches position x2𝑥2 (where x2>x1𝑥2>𝑥1), how much work Wp𝑊p has the person done on the box?Assume that the box reaches x2𝑥2 after the person has accelerated it from rest to speed v1𝑣1.Express the work in terms of m𝑚, v0𝑣0, x1𝑥1, x2𝑥2, and v1𝑣1.View Available Hint(s)for Part BActivate to select the appropriates template from the following choices. Operate up and down arrow for selection and press enter to choose the input value typeActivate to select the appropriates symbol from the following choices. Operate up and down arrow for selection and press enter to choose the input value typeWp𝑊p =
Question
A box of mass m𝑚 is sliding along a horizontal surface.Part AThe box leaves position x=0𝑥=0 with speed v0𝑣0. The box is slowed by a constant frictional force until it comes to rest at position x=x1𝑥=𝑥1.Find Ff𝐹f, the magnitude of the average frictional force that acts on the box. (Since you don't know the coefficient of friction, don't include it in your answer.)Express the frictional force in terms of m𝑚, v0𝑣0, and x1𝑥1.View Available Hint(s)for Part AActivate to select the appropriates template from the following choices. Operate up and down arrow for selection and press enter to choose the input value typeActivate to select the appropriates symbol from the following choices. Operate up and down arrow for selection and press enter to choose the input value typeFf𝐹f =nothingSubmitPart BAfter the box comes to rest at position x1𝑥1, a person starts pushing the box, giving it a speed v1𝑣1.When the box reaches position x2𝑥2 (where x2>x1𝑥2>𝑥1), how much work Wp𝑊p has the person done on the box?Assume that the box reaches x2𝑥2 after the person has accelerated it from rest to speed v1𝑣1.Express the work in terms of m𝑚, v0𝑣0, x1𝑥1, x2𝑥2, and v1𝑣1.View Available Hint(s)for Part BActivate to select the appropriates template from the following choices. Operate up and down arrow for selection and press enter to choose the input value typeActivate to select the appropriates symbol from the following choices. Operate up and down arrow for selection and press enter to choose the input value typeWp𝑊p =
Solution
Part A:
The work done by the frictional force is equal to the change in kinetic energy of the box. The initial kinetic energy of the box is (1/2)mv0^2 and the final kinetic energy is 0 (since the box comes to rest). Therefore, the work done by the frictional force is:
Wf = ΔK = Kf - Ki = 0 - (1/2)mv0^2 = -(1/2)mv0^2
The work done by a force is also equal to the force times the distance over which it acts. Therefore, we can set up the equation:
Wf = Ff * x1
Substituting the expression for Wf from above, we get:
-(1/2)mv0^2 = Ff * x1
Solving for Ff, we find:
Ff = -(1/2)mv0^2 / x1
The negative sign indicates that the frictional force is acting in the opposite direction of the motion of the box.
Part B:
The work done by the person is equal to the change in kinetic energy of the box. The initial kinetic energy of the box is 0 (since the box is at rest) and the final kinetic energy is (1/2)mv1^2. Therefore, the work done by the person is:
Wp = ΔK = Kf - Ki = (1/2)mv1^2 - 0 = (1/2)mv1^2
Note that the distance over which the person pushes the box (x2 - x1) does not affect the amount of work done, since the work done is only dependent on the change in kinetic energy of the box.
Similar Questions
A 15 kg box is being pulled with a force of 45 N and is sliding with an acceleration of 2 m/s2. Find the coefficient of kinetic friction? (
A small box of mass m is placed on a top of a larger box of mass 2m as shown in the diagram. When a force F is applied to the large box, both boxes accelerate to the right with the same acceleration. If the coefficient of friction between all surfaces is μ what would be the force accelerating the smaller mass?
This image shows a box on the ground, the normal force acting on the box, and the force being applied to the box. The box is accelerating under the applied force. If the kinetic friction coefficient between the box and the ground is 0.60, what is the friction force acting on the box?Select one:7.40 N to the right.highlight_off9.00 N to the right.highlight_off32.0 N to the right.highlight_off24.6 N to the right.highlight_off
A box with weight 55 N is on a rough horizontal surface. An external force F is applied horizontally to the box. A normal force and a friction force are also present, denoted by n and f.A force diagram, showing the four forces that act on the box, is shown below:When force F equals 2.0 N, the box is in motion at constant velocity.What is the acceleration of the box when force F equals 2.6 N?Express your answer in m/s2.
A 10.0kg box being pushed at constant velocity against a 3.0N[left] force offriction.a. Draw free body diagram (2 marks)b. Find the coefficient of friction. State if it is static or kinetic. (2 marks)
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.