Two horses pull a large crate of negligible mass at constant speed across a barn

Question

Two horses pull a large crate of negligible mass at constant speed across a barn floor by means of two light steel cables. A large box of mass 250 kg sits on the crate (see the figure below, which shows a top-down view of the crate and box). As the horses pull, the cables are parallel to the horizontal floor. The coefficient of friction between the crate and the barn floor is 0.25.

(a) What is the work done by each horse if the box is moved a distance of 29 m? kJ

(b) What is the tension in each cable if the angle between each and the direction the crate moves is 12

Explanation / Answer

A.) Work is defined two ways, either as the total potential and kinetic energy of the system, or as the force times the distance which the force is applied. Since there is no measurable potential energy or kinetic energy, we will say that W = F*d*Cos(theta). Since theta = 0 (the bar is parallel to the floor), Cos(theta) = 1, so W = F*d = m*a*d = m*-g*mu*d. From here, we plug-n-chug to determine the answer: W = (250kg)*(9.81m/s^2)*(0.25)(29m) = 1776285J B.) Tension is defined as the total force applied to the cable. Since there are two cables. Since our W = F*d = 18491.85J, our F = 17762.85J/29m = 637.6N. Dividing by Cos(13), we get that our tension is going to be T = 654.4N in total. Since there are two cables, this tension is going to be divided by 2, so our tension in each cable is 327.2N. So the tension in each cable is going to be 1308.85N.

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