Problem 86 - University Physics with Modern Physics

Question

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.

Step-by-Step Solution

Answer

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 1: 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. \(F_{hor} = F_{father} * \cos(\theta) = 2.4*10^2 *\cos(20^{\circ}) =240*\cos(20^{\circ}) \approx 225.15 \mathrm{~N}\).

Step 2: 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, \(g = 9.81 \mathrm{~m/s^2}\). \(F_{normal} = m * g = 85.0 * 9.81 \approx 833.85 \mathrm{~N}\).

Step 3: Calculate the friction force.

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

Step 4: 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, \(W_{father} = F_{hor} * d * \cos(\alpha) = 225.15 * 8.00 * \cos(0^\circ) = 225.15 * 8 \approx 1801.2 \mathrm{~J}\).

Step 5: 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 \(180^\circ\). \(W_{friction} = F_{friction} * d * \cos(\alpha) = 166.77 * 8.00 * \cos(180^\circ) = 166.77 * 8 * (-1) \approx -1334.16 \mathrm{~J}\).

Step 6: 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 \(467.04 \mathrm{~J}\).

Key Concepts

Kinetic Friction

Kinetic friction occurs when two objects slide against each other. In this problem, the sled and the surface create kinetic friction as the sled is pulled across it. Friction always opposes motion, so it acts in the opposite direction of the sled. The force of kinetic friction is calculated using the formula:\[ F_{friction} = \mu \times F_{normal} \]where \( \mu \) is the coefficient of kinetic friction and \( F_{normal} \) is the normal force. Here, the coefficient is given as 0.200, and the normal force is determined by the weight of the sled and the daughter. Kinetic friction plays a major role in determining the work done by friction. Since friction opposes the movement, the work is negative, illustrating that friction takes energy out of the system. Understanding kinetic friction helps in calculating the total energy required to overcome this force and move the sled.

Horizontal Force

The horizontal force in physics is the force exerted parallel to the surface on which the object is moving. For this problem, the father's pulling force has a horizontal component because it is applied using a rope at an angle. To find this component, we use the formula:\[ F_{hor} = F_{father} \times \cos(\theta) \]where \( F_{father} \) is the force exerted by the father, and \( \theta \) is the angle of the rope with the horizontal. By solving this, we know how much force is actually moving the sled forward. This component is critical for calculating the work done, since work is the product of the force and the distance in the direction of the force.

Angle of Force

When a force is applied at an angle, only a portion of it contributes to the horizontal and vertical motion. In this exercise, the father's pulling force makes a \(20^{\circ}\) angle with the horizontal. The angle influences both the effective horizontal force and how much the vertical component affects the normal force.Understanding the angle of application is crucial because:

• It determines how effectively the force contributes to moving the object horizontally.
• The vertical component affects the normal force, which in turn influences the frictional force.

By considering the angle, we can decompose the pulling force into horizontal and vertical components, making our calculations more accurate and precise.

Normal Force

The normal force is the support force exerted on an object in contact with a surface, acting perpendicular to the surface. In this scenario, the normal force is crucial as it influences the magnitude of the frictional force.Since there is no vertical motion, the normal force balances the gravitational force acting on the sled.\[ F_{normal} = m \times g \]where \( m \) is the mass of the sled including the daughter, and \( g \) is the acceleration due to gravity (approximately \(9.81 \ \mathrm{m/s^2}\)). However, since the force is applied at an angle, only part of the gravitational force contributes to the normal force after accounting for the vertical component of the applied force. Understanding the normal force allows us to accurately calculate the kinetic friction which impacts the work required to move the sled.