Help with Physics Problem Question

[flisally]

Hi guys, the following question is in Chad's videos physics Work Quiz:

Two 50kg sleds connected by a rope are pulled horizontally at constant velocity by four dogs across ice with negligible friction so that the tension in the rope is 200N. What is the work performed by the dogs over 100m?

The answer says:
W = (Fcosθ)d

W = Fd since the force and displacement are in the same direction


The displacement is given as 100m but to solve for the work performed by the dogs their pulling force must be determined.


The acceleration of the box on the left can be calculated from Newton's 2nd law applied in the horizontal direction:

SF = ma

200N = (50kg)a

a = 4m/s2


As the two blocks are tethered to each other and move together this will also be the acceleration of the box on the right, and we can use Newton's 2nd law applied in the horizontal direction to the box on the right to calculate the dogs' pulling force.

SF = ma

Fdogs - 200N = (50kg)(4m/s2)

Fdogs = 400N


W = Fd

W = (400N)(100m)

W = 40,000J = 40kJ


But in the question, it says that the two sleds travel at constant velocity, so wouldn't acceleration be zero? Can someone help explain this to me? Thanks so much in advance.

 

[lwergod]

are you sure that's what the question says? I think it should be saying, the dogs accelerate at a constant velocity...

if the dogs are moving at constant velocity on friction-less surface, then there shouldn't be any tension in the rope.

btw, you should be posting this in the mcat section of the forums.

 

[flisally]

are you sure that's what the question says? I think it should be saying, the dogs accelerate at a constant velocity...

if the dogs are moving at constant velocity on friction-less surface, then there shouldn't be any tension in the rope.

btw, you should be posting this in the mcat section of the forums.

I just copied and pasted the question so I think the question may have a typo then.

Sorry I posted in the wrong section!

Thanks for your help.