Question on Work done by horse pulling a wagon.

[Daaniyaal]

Homework Statement

How much work did a horse do that pulled a 200-kg wagon 50 km without acceleration along a level road if the effective coefficient of friction was 0.060?

Homework Equations

W=F_net * d

Force of friction= µmg

The Attempt at a Solution

W=200*9.8-(200*9.8*0.060)*50,000
=92120000


Correct answer is supposed to be 5.9*10⁶

 

[Chestermiller]

Your equation doesn't make sense. Where did that first term come from? The correct answer should just be the second term with a plus sign. If you recalculate the expression you gave, you will find that you evaluated it incorrectly.

Chet

 

[tiny-tim]

Hi Daaniyaal!

W=200*9.8-(200*9.8*0.060)*50,000

work done is force "dot" displacement …

what is the dot product of the vertical weight with the horizontal displacement?

 

[Daaniyaal]

I guess I unnecessarily put in the force needed to move the wagon without friction opposing the movement. Oopsies



I feel embarrassed >.>

Also, thank you :)

 

[Nickie]

I would like to clarify that the correct unit for work is Joules (J) and not Newtons (N). The correct formula for work is W = F*d, where F is the force applied and d is the distance traveled. In this case, the force applied by the horse is the net force, which is the difference between the force of gravity (200*9.8 N) and the force of friction (0.060*200*9.8 N). Therefore, the work done by the horse can be calculated as:

W = (200*9.8 N - 0.060*200*9.8 N) * 50,000 m = 5.9*10^6 J

This is equivalent to the given answer of 5.9*10^6 J or 5.9 million Joules. It is important to use the correct units and formula when solving physics problems to ensure accurate results.