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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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Most popular questions from this chapter

A package is dropped on a horizontal conveyor belt. The mass of the package is \(m,\) the speed of the conveyor belt is \(v\), and the coefficient of kinetic friction between the package and the belt is \(\mu_{\mathrm{k}}\) a) How long does it take for the package to stop sliding on the belt? b) What is the package's displacement during this time? c) What is the energy dissipated by friction? d) What is the total work supplied by the system?

A block of mass \(5.0 \mathrm{~kg}\) slides without friction at a speed of \(8.0 \mathrm{~m} / \mathrm{s}\) on a horizontal table surface until it strikes and sticks to a mass of \(4.0 \mathrm{~kg}\) attached to a horizontal spring (with spring constant of \(k=2000.0 \mathrm{~N} / \mathrm{m}\) ), which in turn is attached to a wall. How far is the spring compressed before the masses come to rest? a) \(0.40 \mathrm{~m}\) b) \(0.54 \mathrm{~m}\) c) \(0.30 \mathrm{~m}\) d) \(0.020 \mathrm{~m}\) e) \(0.67 \mathrm{~m}\)

A cannonball of mass \(5.99 \mathrm{~kg}\) is shot from a cannon at an angle of \(50.21^{\circ}\) relative to the horizontal and with an initial speed of \(52.61 \mathrm{~m} / \mathrm{s}\). As the cannonball reaches the highest point of its trajectory, what is the gain in its potential energy relative to the point from which it was shot?

The greenskeepers of golf courses use a stimpmeter to determine how "fast" their greens are. A stimpmeter is a straight aluminum bar with a V-shaped groove on which a golf ball can roll. It is designed to release the golf ball once the angle of the bar with the ground reaches a value of \(\theta=20.0^{\circ} .\) The golf ball \((\) mass \(=1.62 \mathrm{oz}=0.0459 \mathrm{~kg})\) rolls 30.0 in down the bar and then continues to roll along the green for several feet. This distance is called the "reading." The test is done on a level part of the green, and stimpmeter readings between 7 and \(12 \mathrm{ft}\) are considered acceptable. For a stimpmeter reading of \(11.1 \mathrm{ft},\) what is the coefficient of friction between the ball and the green? (The ball is rolling and not sliding, as we usually assume when considering friction, but this does not change the result in this case.)

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