## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 6 - Problems - Page 227: 8

#### Answer

To pull the cart, the horse needs to eat 433 grams of oats.

#### Work Step by Step

We can find the work $W_h$ done by the horse on the cart: $Work = \Delta KE$ $W_h-F_f~d = \frac{1}{2}mv^2$ $W_h = \frac{1}{2}mv^2+F_f~d$ $W_h = \frac{1}{2}(250~kg)(0.38~m/s)^2+(260~N)(1500~m)$ $W_h = 3.9\times 10^5~J$ This work is only 10% of the energy that the horse gets from eating oats. Therefore, to pull the cart, the horse needs to eat enough oats to get $3.9\times 10^6~J$ of energy. We can find the number of grams of oats that the horse needs to eat: $\frac{3.9\times 10^6~J}{9000~J/gram} = 433~grams$ To pull the cart, the horse needs to eat 433 grams of oats.

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