Friction and Work

[HopefulOncoDoc]

A car traveling on a dry road at 10 m/s comes to a stop (uk = 0.5). If the same car were instead traveling on a wet road (uk = 0.2), how does the work required to stop the car on the wet road, Wwet, compare to the work required to stop the car on the dry road, Wdry?

A. Wdry = 2 Wwet
B. Wdry = Square Root(2)Wwet
C. Wdry = Wwet
D. Wwer = 2 Wdry

The answer is C. Could someone please explain this one?

What can be said about the speed of an object if the net work done on that object is zero?

A. The final and initial speeds are the same.
B. The speed of the object was constant.
C. The final speed was larger than the initial speed.
D. The initial speed was larger than the final speed.

The answer is A but choice B sounds like a better choice? Could someone please explain this one as well?

Thanks a lot.

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[ThirdEye]

A car traveling on a dry road at 10 m/s comes to a stop (uk = 0.5). If the same car were instead traveling on a wet road (uk = 0.2), how does the work required to stop the car on the wet road, Wwet, compare to the work required to stop the car on the dry road, Wdry?

A. Wdry = 2 Wwet
B. Wdry = Square Root(2)Wwet
C. Wdry = Wwet
D. Wwer = 2 Wdry

The answer is C. Could someone please explain this one?

What can be said about the speed of an object if the net work done on that object is zero?

A. The final and initial speeds are the same.
B. The speed of the object was constant.
C. The final speed was larger than the initial speed.
D. The initial speed was larger than the final speed.

The answer is A but choice B sounds like a better choice? Could someone please explain this one as well?

Thanks a lot.

1. They do the same work, it just takes longer for the wet surface. Work is a transfer of energy and the car has the same energy in both situations.

2. Net work of 0 means that work is either cancled out or doesn't exist. If work is done in one direction, and later on equal work is done in the opposite direction, then the initial and final speeds are equal but there was a change in speed between initial and final.

 

[Hemichordate]

Net work of 0 means that work is either cancled out or doesn't exist. If work is done in one direction, and later on equal work is done in the opposite direction, then the initial and final speeds are equal but there was a change in speed between initial and final.

But couldn't you also say that net work = 0 ---> net force = 0 -----> acceleration = 0 -------> constant velocity?

 

[Hemichordate]

bump

 

[HopefulOncoDoc]

1. They do the same work, it just takes longer for the wet surface. Work is a transfer of energy and the car has the same energy in both situations.

So W on by friction = -fkineticΔx. In the Wdry case, Δx would be shorter since it has a higher coefficent of kinetic friction but in the Wwet case, Δx would be higher because it has a shorter coefficient of kinetic friction and thus they essentially do the same amount of W, would this be correct?

 

[Omni]

Frictional work in a sense can also be seen as heat. In both cases, the car is going at 10 m/s hence KE for both is the same. Friction must take away the same amount of energy in either pavement. in the wet pavement, this is done on a longer surface, whereas on the dry, it is done on a shorter distance.
Yeah, you're right hopeful

 

[ThirdEye]

But couldn't you also say that net work = 0 ---> net force = 0 -----> acceleration = 0 -------> constant velocity?

It could be the case ist's just not necessarily the case. If I move an object to the right and then back to it's original position, the net force is zero but there was an acceleration and velocity wasn't constant. Initial and final velocity are equal however.