What is the work done by a horse pulling a cart?

[walking]

The answers for author's solution differ slightly from mine when I convert his answers to SI. I think this is only a conversion issue as the differences are small but I wanted to confirm if my solution is correct:

d=rt (distance = rate x time) so (12 min)(6.2mi/h)(60s/1min)(0.3048m/s)(0.6816mi/h). Then $$W=(F\cos \theta)d=(45lb\cos 27)d$$ etc. I end up getting 355940J wjereas author gets 332190J (he doesn't give answer in J since he works with feet, but after converting it to J using accurate conversion rates this is what I got).

 

[kuruman]

I got 244,921 lb*ft which converts to 332,068 Joules. My answer differs from the author's by 0.04%. Yours differs by 7.1%. It looks like it's more accurate to work with feet and do one final conversion in the end.

What did you get for part (b)?

 

[BvU]

It doesn't say 45 lb, it says 42 lb.

356 or 332 is too far off for conversion accuracy

Your

(12 min)(6.2mi/h)(60s/1min)(0.3048m/s)(0.6816mi/h)

has the wrong dimension.
What is (0.3048 m/s)(0.6816 mph) ?

I get 332.1904 kW $\qquad$

Note that providing more than two digits is questionable: it suggests more accuracy than actually justified

$\$

 

[kuruman]

It doesn't say 45 lb, it says 42 lb.

And the discrepancy between 45 and 42 is 7.1%. I didn't notice the mistaken input.

 

[walking]

Right, so I made the mistake of using 45lb instead of 42. When I use 42 I get $$(42lb\cos 27)(12min)(6.2mi/h)\left( \frac{10^{-5}N}{2.248\times 10^{-6}lb}\right)\left(\frac{60s}{1min}\right)\left(\frac{0.3048m/s}{0.6818mi/h}\right)=332210=3.32\times 10^5 J$$