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A student decides to move a box of books into her dormitory room by pulling on a rope attached to the box. She pulls with a force of \(80.0 \mathrm{~N}\) at an angle of \(25.0^{\circ}\) above the horizontal. The box has a mass of \(25.0 \mathrm{~kg}\), and the coefficient of kinetic friction between box and floor is \(0.300 .\) (a) Find the acceleration of the box. (b) The student now starts moving the box up a \(10.0^{\circ}\) incline, keeping her \(80.0 \mathrm{~N}\) force directed at \(25.0^{\circ}\) above the line of the incline. If the coefficient of friction is unchanged, what is the new acceleration of the box?

Short Answer

Expert verified
The acceleration of the box when the student is pulling the box on the flat surface is \(1.9 m/s^2\), and the acceleration when pulling the box up the incline is \(0.03 m/s^2\).

Step by step solution

01

Determine the effective force

First, begin by determining the force with which the student is pulling the box. This force can be broken down into horizontal and vertical components. The horizontal component \(F_{h1}\) of the force can be computed using the cos function as follows: \(F_{h1} = F \times \cos(25)\) where \(F = 80N\). The vertical component \(F_{v1}\) can be calculated using the sin function as such: \(F_{v1} = F \times \sin(25)\). Gravity also acts on the box: \(F_g = m \times g\) where \(m = 25kg\) and \(g = 9.8m/s^2 \). Therefore the net force in horizontal direction will be \(F_{net1} = F_{h1} - (F_{k1} + F_g)\), where \(F_{k1}\) is the kinetic friction = \(\mu \times Normal~Force = .3 \times (m \times g - F_{v1})\).
02

Calculate Acceleration

The acceleration of the box can then be determined using Newton’s second law which states that \(Force = mass \times acceleration\). Therefore, \(a = F_{net1}/m\).
03

Repeat for the inclined surface

For the inclined plane we have similar force components, however, the force due to gravity is now split into two components - one along the inclined surface and the other perpendicular to the surface. Also, the student's force is now considered as inclined at an angle to the incline. By repeating the above steps to compute forces and then dividing by the mass of the box, the new acceleration can be found.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Force Components
In physics, understanding the components of force is crucial when analyzing motion, particularly in scenarios that involve forces applied at angles. When an object is pulled at an angle, the force exerted has both horizontal and vertical components.

For instance, if a student pulls a rope attached to a box at an angle, as in our exercise, we must break this force into its horizontal component (parallel to the surface) and its vertical component (perpendicular to the surface). Mathematically, we use trigonometric functions to do this: the horizontal component is found using the cosine of the angle (\(F_{h} = F \cdot \cos(\theta)\)) and the vertical component using the sine of the angle (\(F_{v} = F \cdot \sin(\theta)\)) where \(F\) is the magnitude of the applied force and \(\theta\) is the angle of application.

These components are essential in calculating the net force that will affect an object's acceleration; in the presence of friction, it is especially important as frictional force opposes the motion along the surface, directly subtracting from the horizontal component of the applied force.
Newton's Second Law
Newton's second law of motion is a cornerstone in physics, particularly in understanding motion and force. It states that the acceleration of an object is directly proportional to the net force acting upon it and inversely proportional to its mass (\(a = \frac{F_{net}}{m}\)). Simply put, larger forces cause greater accelerations, while larger masses resist acceleration more strongly.

In application, after you calculate the net force by taking into account all the forces acting on an object, including friction and component forces, you can determine the acceleration using this law. For the box in the exercise, using the force components, we subtract the force of friction from the horizontal component of the pulling force to calculate the net force. Dividing this by the mass of the box gives the acceleration. This direct relationship helps explain how varying either the force applied or the mass of the object will change the acceleration - an important consideration when analyzing the motion of objects in a variety of contexts.
Inclined Plane Physics
An inclined plane simplifies lifting heavy objects by increasing the distance over which the force is applied, but it introduces complexity in force analysis due to gravity acting at an angle relative to the surface. When an object moves on an incline, gravity has components both parallel and perpendicular to the surface. The parallel component (\(F_{g\parallel}\)) causes the object to accelerate down the plane, while the perpendicular component (\(F_{g\perp}\)) contributes to the normal force, which in turn affects friction.

In the exercise, with the box moving up an incline, we repeat the force component analysis considering the incline angle. Both the student's pulling force and gravity must be resolved into parallel and perpendicular components relative to the incline. The friction force still acts to oppose the motion along the incline, proportional to the normal force. These modifications in the force calculations result in a different net force when compared to the horizontal scenario, thus affecting the acceleration of the box differently. Understanding the physics of inclined planes is vital when calculating forces and predicting the motion of objects on ramps or slopes.

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Most popular questions from this chapter

A sled weighing \(60.0 \mathrm{~N}\) is pulled horizontally across snow so that the coefficient of kinetic friction between sled and snow is \(0.100, \overline{\mathrm{A}}\) penguin weighing \(70.0 \mathrm{~N}\) rides on the sled, as in Figtire \(\mathrm{P} 4.78\). If the coefficient of static friction between penguin and sled is \(0.700\), find the maximum horizontal force that can be exerted on the sled before the penguin begins to slide off.

The air exerts a forward force of \(10 \mathrm{~N}\) on the propeller of a \(0.20-\mathrm{kg}\) model airplane. If the plane accelerates forward at \(2.0 \mathrm{~m} / \mathrm{s}^{2}\), what is the magnitude of the resistive force exerted by the air on the airplane?

A shopper in a supermarket pushes a loaded cart with a horizontal force of \(10 \mathrm{~N}\). The cart has a mass of \(30 \mathrm{~kg}\). (a) How far will it move in \(3.0\) s, starting from rest? (Ignore friction.) (b) How far will it move in \(3.0 \mathrm{~s}\) if the shopper places his \(30-\mathrm{N}\) child in the cart before he begins to push it?

A \(5.0-\mathrm{kg}\) bucket of water is raised from a well by a rope. If the upward acceleration of the bucket is \(3.0 \mathrm{~m} / \mathrm{s}^{2}\), find the force exerted by the rope on the bucket.

An inquisitive physics student, wishing to combine pleasure with scientific inquiry, rides on a roller coaster sitting on a bathroom scale. (Do not try this yourself on a roller coaster that forbids loose, heavy packages.) The bottom of the seat in the roller-coaster car is in a plane parallel Lo the track. The seat has a perpendicular back and a seat belt that fits around the student's chest in a plane parallel to the bottom of the seat. The student lifts his feet from the floor so that the scale reads his weight, \(200 \mathrm{lb}\), when the car is horizontal. At one point during the ride, the car zooms with negligible friction down a straight slope inclined at \(30.0^{\circ}\) below the horizontal. What does the scale read at that point?

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