Question

Starting from rest, a horse pulls a 263−kg cart for a distance of 1.6 km. It reaches a speed of 0.38 m/s by the time it has walked 50.0 m and then walks at constant speed. The frictional force on the rolling cart is a constant 253 N. Each gram of oats the horse eats releases 9 kJ of energy; 10 % of this energy can go into the work the horse must do to pull the cart. How many grams of oats must the horse eat to pull the cart?

Answer 1

Answer:

$m=449.79 g$

Explanation:

The work done on the cart by the horse is found by:

$W= K_E+F_K$

$W=\frac{1}{2}*m*v^2+f_k*d$

replacing:

$E=9kJ$, $f_k=253N$, $d=1600.0m$, $v=0.38m/s$, $m=263kg$

$W=\frac{1}{2}*263kg*(0.38m/s)^2+253N*1600m$

$W=404818.98 J$

$W=404.818x10^3 J$

10% E= 900J

$W=E*m$

$m=\frac{W}{E}=\frac{404.81x10^3J}{900J/g}$

$m=449.79 g$

Answer 2

This question involves the concepts of work done and mass. It can be solved by calculating the total work done by the horse.

The horse will have to eat "449.83 grams" of oats to pull the cart.

First, we will find the work done by the horse, using the law of conservation of energy.

$W = K.E+W_f$

where,

W = Work Done

K.E = kinetic energy

$W_f$ = work done by friction

Therefore,

$W = \frac{1}{2}mv^2+fd$

where,

m = mass of cart = 263 kg

v = speed of cart = 0.38 m/s

f = frictional force = 253 N

d = distance traveled = 1.6 km

Therefore,

$W = \frac{1}{2}(263\ kg)(0.38\ m/s)^2+(253\ N)(1600\ m)\\\\W = 49.97\ J + 404800\ J\\W = 404849.97\ J = 404.85\ KJ$

It is given in the question that the energy gained by horse for doing work per gram of oats is:

E = 10% of 9 KJ/g

E = (0.1)(9 KJ/g) = 0.9 KJ/g

Hence, the mass of oats required by the horse to pull the cart can be calculated as follows:

$Mass\ of\ Oats=\frac{W}{E}\\\\Mass\ of\ Oats=\frac{404.85\ KJ}{0.9\ KJ/g}$

Mass of Oats = 449.83 grams

Learn more about work done here:

brainly.com/question/13662169?referrer=searchResults

The attached picture explains the work done formula.