/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 46 A \(1200-\mathrm{kg}\) car coast... [FREE SOLUTION] | 91Ó°ÊÓ

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A \(1200-\mathrm{kg}\) car coasts from rest down a driveway that is inclined \(20^{\circ}\) to the horizontal and is \(15 \mathrm{~m}\) long. How fast is the car going at the end of the driveway if \((a)\) friction is negligible and \((b)\) a friction force of \(3000 \mathrm{~N}\) opposes the motion?

Short Answer

Expert verified
Without friction, the car's speed is approximately 12.72 m/s; with friction, it is approximately 7.54 m/s.

Step by step solution

01

Identify the given values and define variables

The mass of the car is \( m = 1200 \mathrm{~kg} \), the incline angle is \( \theta = 20^{\circ} \), and the length of the driveway is \( s = 15 \mathrm{~m} \). We need to find the final velocity \( v \) of the car.
02

Calculate gravitational potential energy loss

The loss in potential energy when the car travels down the incline is given by \( \Delta U = mgh \), where \( h = s \sin \theta \). Calculate \( h \) as follows: \( h = 15 \sin(20^{\circ}) \). Then calculate \( \Delta U = 1200 \times 9.81 \times h \).
03

Determine kinetic energy and final velocity without friction

With negligible friction, the entire potential energy loss converts to kinetic energy (\( KE \)), where \( KE = \frac{1}{2}mv^2 \). Set \( \Delta U = \frac{1}{2}mv^2 \) and solve for \( v \) as \( v = \sqrt{2\Delta U/m} \).
04

Account for frictional force in energy balance

When the friction force \( f = 3000 \mathrm{~N} \) opposes motion, the work done by friction is \( W_f = f \cdot s \). Subtract the work done against friction from the potential energy:\( \Delta U_{net} = \Delta U - W_f \).
05

Calculate final velocity with friction

Use the net potential energy (from step 4) to find the kinetic energy and hence final velocity: \( \Delta U_{net} = \frac{1}{2}mv^2 \) and solve for \( v \) with the new \( \Delta U_{net} \).
06

Conclusion with calculated velocities

Compute the specific velocity values using the formulas derived in the previous steps. Without friction, \( v \) is calculated from the full gravitational potential energy. With friction, \( v \) is calculated using the reduced energy due to work done against friction.

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Key Concepts

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

Kinetic Energy
Kinetic energy is the energy of motion. For an object like the car in our example, it is calculated as \( KE = \frac{1}{2}mv^2 \), where \( m \) is the mass and \( v \) is the velocity. This formula shows that kinetic energy relies heavily on the velocity. In fact, if the velocity doubles, the kinetic energy becomes four times greater.

In the absence of friction, any potential energy lost as an object moves down an inclined plane is converted into kinetic energy. This is why no work done against friction allows all the potential energy to become kinetic energy, resulting in a higher final speed. Understanding how kinetic energy changes in different scenarios helps us predict the motion of objects in the real world.
Gravitational Potential Energy
Gravitational potential energy (GPE) is the energy stored due to an object's position relative to the Earth. It is calculated with the formula \( U = mgh \), where \( m \) is mass, \( g \) is the acceleration due to gravity, and \( h \) is the height above the reference point.

In the context of an inclined plane, this height is determined by the vertical component of the length of the slope. The car in the example has a potential energy related to its height before it starts moving down the inclined driveway. The energy change due to a drop in height translates directly into kinetic energy if friction is negligible, illustrating the conservation of energy. This concept highlights the seamless conversion of energy from one form to another.
Inclined Plane
An inclined plane is a flat surface that is tilted at an angle compared to the horizontal. This concept reduces the required force over a distance, but doesn't change the total work needed to move an object from the bottom to the top of the slope. In solving physics problems involving inclined planes, it's crucial to analyze the components of forces acting along the incline and perpendicular to it.

With the example of the car, the slope of the driveway is an inclined plane. The angle of incline influences the gravitational component that pulls the car downward. Knowledge of how inclined planes work aids in understanding mechanics in a spatial context.
Frictional Force
Frictional force is an opposing force that acts against the motion of an object. It plays a crucial role in physics problems by reducing the net energy available to be converted into motion. In our exercise, a frictional force of \( 3000 \mathrm{~N} \) opposes the motion of the car, influencing its final velocity.

To calculate the effect of friction, we need to determine the work done by the frictional force, which is calculated as \( W_f = f \cdot s \), with \( f \) being the frictional force and \( s \) being the distance moved along the plane. This work done by friction reduces the net potential energy, resulting in less energy to convert into kinetic energy, and subsequently a lower speed. Friction is everywhere, and understanding it helps us account for energy losses and real-world motion dynamics.

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Most popular questions from this chapter

How much work is done on an object by the force that supports it as the object is lowered at a constant speed through a vertical distance \(h\) ? How much work does the gravitational force on the object do in this same process? The supporting force is \(m g\), where \(m\) is the mass of the object. It is directed upward while the displacement is downward. Hence, the work it does is negative: $$ F \cos \theta=(m g)(h)\left(\cos 180^{\circ}\right)=-m g h $$ The force of gravity acting on the object is also \(m g\), but it is directed downward in the same direction as the displacement. The work done on the object by the force of gravity is therefore positive: $$ F \operatorname{scos} \theta=(m g)(h)\left(\cos 0^{\circ}\right)=m g h $$

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A \(10.0\) -kg flowerpot falls off a windowsill \(30.0 \mathrm{~m}\) above the street. In falling to the ground, how much work is done on the pot by the gravitational interaction?

A proton \(\left(m=1.67 \times 10^{-27} \mathrm{~kg}\right)\) that has a speed of \(5.0 \times 10^{6} \mathrm{~m} / \mathrm{s}\) passes through a metal film of thickness \(0.010 \mathrm{~mm}\) and emerges with a speed of \(2.0 \times 10^{6} \mathrm{~m} / \mathrm{s}\). How large an average force opposed its motion through the film?

Compute the power output of a machine that lifts a 500 -kg crate through a height of \(20.0 \mathrm{~m}\) in a time of \(60.0 \mathrm{~s}\).

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