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Chapter 6: Problem 86

A father exerts a \(2.40 \cdot 10^{2} \mathrm{~N}\) force to pull a sled with his daughter on it (combined mass of \(85.0 \mathrm{~kg}\) ) across a horizontal surface. The rope with which he pulls the sled makes an angle of \(20.0^{\circ}\) with the horizontal. The coefficient of kinetic friction is \(0.200,\) and the sled moves a distance of \(8.00 \mathrm{~m}\). Find a) the work done by the father, b) the work done by the friction force, and c) the total work done by all the forces.

Short Answer

Expert verified
The work done by the father is 1801.2 Joules, the work done by the friction force is -1334.16 Joules, and the total work done by all the forces is 467.04 Joules.

Step by step solution

01

Calculate the horizontal component of the father's force.

We can calculate the horizontal component of the father's force by using the given force and angle. \(F_{hor} = F_{father} * \cos(\theta) = 2.4*10^2 *\cos(20^{\circ}) =240*\cos(20^{\circ}) \approx 225.15 \mathrm{~N}\).
02

Calculate the normal force.

Since there is no vertical motion, the normal force is equal to the weight of the sled and daughter, which is the product of their mass and the gravitational acceleration, \(g = 9.81 \mathrm{~m/s^2}\). \(F_{normal} = m * g = 85.0 * 9.81 \approx 833.85 \mathrm{~N}\).
03

Calculate the friction force.

Now we can determine the friction force acting on the sled. \(F_{friction} = \mu * F_{normal} = 0.200 * 833.85 \approx 166.77 \mathrm{~N}\).
04

Calculate the work done by the father.

The work done by the father can be calculated by considering only the horizontal component of the force exerted by him. Since the horizontal force and displacement are in the same direction, the angle \(\alpha\) is \(0^\circ\). So, \(W_{father} = F_{hor} * d * \cos(\alpha) = 225.15 * 8.00 * \cos(0^\circ) = 225.15 * 8 \approx 1801.2 \mathrm{~J}\).
05

Calculate the work done by the friction force.

The work done by the friction force is in the opposite direction to the displacement, so the angle \(\alpha\) is \(180^\circ\). \(W_{friction} = F_{friction} * d * \cos(\alpha) = 166.77 * 8.00 * \cos(180^\circ) = 166.77 * 8 * (-1) \approx -1334.16 \mathrm{~J}\).
06

Calculate the total work done by all the forces.

The total work done by all the forces is the sum of the work done by the father and the work done by the friction force: \(W_{total} = W_{father} + W_{friction} = 1801.2 - 1334.16 \approx 467.04 \mathrm{~J}\). So, the work done by the father is \(1801.2 \mathrm{~J}\), the work done by the friction force is \(-1334.16 \mathrm{~J}\), and the total work done by all the forces is \(467.04 \mathrm{~J}\).

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