How Is Work Calculated in a Friction-Involved Scenario?

[yue_]

Homework Statement

There is a 2450 kg car being pushed by a group of students. The car is pushed a distance of 12 m in 8.95 seconds. The coefficient of friction is .450.
a) What is the net work being done?
b) Work done by street?
c) Work done by students?

Homework Equations

Wnet=Kf-K0
Wf=-Ff*d
d=[(v0+vf)t]/2
KE=[mv^2]/2

The Attempt at a Solution

For a, Wnet=Kf-K0, and K0=0.
To find Kf, I started to find Vf, using the displacement formula
12=[vf*8.95]/2
vf=2.68 sec.
Then I found Kf
Kf=[2450*2.68^2]/2
Since K0=0, Wnet=8798 J (same value as Kf+0)

b asks for work done by street so I assumed that meant Wf.
At this point, I wasn't sure how to find Ff to use in Wf=-Ff*d formula
I thought Ff=μ*Fn
Is Fn=m*g? Or do I multiply m*a, after finding out acceleration?
Since I found Vf in the last step, do I divide that by time to find acceleration?

c is work done by students, which is Wpush/applied force.
Wnet=Wpull-Wf
Do I add Wf to Wnet to find applied force?
At this point, I don't really know what I'm doing.
Can someone give me a few tips on the concepts that should be understood about this problem and explain whatever errors I'm making?

 

[Zula110100100]

Ff is μ*Fn This will give you the force of friction, and you know the distance over which it was acting. And I am not sure what is better, but I personally would have found the acceleration with s = (at^2)/2
12 = a(8.95^2)/2
24 = 80.1025a
a = .2996m/s^2

Fnet = ma = .2996*2450 = 734.02
Wnet = Fnet*s = 734.02*12 = 8808J

Pretty similar answer

 

[yue_]

Ff is μ*Fn This will give you the force of friction, and you know the distance over which it was acting.

Thanks for replying. I did get that acceleration and a very close Wnet (8820 J) when I worked it previously, while using the method you demonstrated.

For Ff, is Fn=m*a? Or m*g? That's what I'm confused about...

 

[JHamm]

Fn is the normal force, which is mg :)

 

[asteriatic]

As a scientist, my response to this content would be:

First, it is important to clarify that this problem deals with the concept of work, which is defined as the product of force and displacement. In this case, the work being done is the net work, which takes into account both the work done by the students pushing the car and the work done by the street acting against the car's motion.

To solve for the net work, we can use the formula Wnet=Kf-K0, where Kf is the final kinetic energy and K0 is the initial kinetic energy. Since the car starts from rest, K0=0 and we only need to find Kf. Using the equation for displacement, we can find the final velocity of the car, which is 2.68 m/s. Then, using the formula for kinetic energy, we can find Kf to be 8798 J.

For part b, we are asked to find the work done by the street, which is also known as the work done by friction (Wf). To solve for Wf, we can use the formula Wf=-Ff*d, where Ff is the force of friction and d is the displacement. The coefficient of friction (μ) is given as 0.450, which is multiplied by the normal force (Fn) to find the force of friction. Fn can be calculated as the product of the mass of the car (2450 kg) and the acceleration due to gravity (9.8 m/s^2). Thus, we can find Wf to be -1188 J.

Lastly, for part c, we are asked to find the work done by the students, which is also known as the work done by the applied force. We can use the formula Wpush= Wnet + Wf to find the work done by the students. This is because the net work takes into account both the work done by the students and the work done by friction. Thus, we can find Wpush to be 8798 J - 1188 J = 7610 J.

In summary, to solve this problem, it is important to understand the concept of work, and how it is affected by forces and displacements. It is also crucial to use the correct formulas and values to accurately solve for the net work, work done by friction, and work done by the students.