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Chapter 6: Problem 40

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: 9.14 m/s; With friction: 6.87 m/s.

Step by step solution

01

Understanding the Problem

We need to find the car's speed at the bottom of a 15 m inclined driveway. The car starts from rest, and the incline is at 20 degrees with the horizontal. We'll consider two scenarios: (a) without friction and (b) with a friction force of 3000 N.
02

Calculate Potential Energy

The potential energy at the top of the driveway is given by the equation: \( PE = m \cdot g \cdot h \), where \( m = 1200 \text{ kg} \), \( g = 9.8 \text{ m/s}^2 \), and \( h \) is the vertical height. To find \( h \), use \( h = 15 \cdot \sin(20^{\circ}) \).
03

Convert Potential Energy to Kinetic Energy (Without Friction)

In scenario (a), all potential energy converts to kinetic energy at the bottom. Set initial kinetic energy to zero since the car starts from rest. Use \( KE = \frac{1}{2} m v^2 \) and solve for \( v \). Hence, \( m \cdot g \cdot h = \frac{1}{2} m v^2 \) leads to \( v = \sqrt{2gh} \).
04

Calculate for Scenario A

Substitute the values: \( v = \sqrt{2 \cdot 9.8 \cdot 15 \cdot \sin(20^{\circ})} \). Calculate this to find the car's speed without friction.
05

Apply Work-Energy Principle (With Friction)

For scenario (b), the work done by friction \( W_f = - f \cdot d \). The work-energy principle: \( KE_{final} = PE_{initial} + W_{friction} \), where \( f = 3000 \text{ N} \) and \( d = 15 \text{ m} \).
06

Solve for Final Speed with Friction

Write out the equation: \( \frac{1}{2} m v^2 = mgh - fd \). Substitute all known values and solve for \( v \) in this scenario. Calculate \( v = \sqrt{2 \left(g \cdot h - \frac{3000 \times 15}{1200}\right)} \).
07

Determine the Numerical Values

Calculate both scenarios: For (a), find \( v = \sqrt{2 \cdot 9.8 \cdot 15 \cdot \sin(20^{\circ})} \). For (b), calculate \( v = \sqrt{2 \left(9.8 \cdot 15 \cdot \sin(20^{\circ}) - \frac{3000 \times 15}{1200}\right)} \).

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Key Concepts

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

Kinetic Energy
Kinetic energy, simply put, is the energy of motion. When something is moving, it has kinetic energy. The formula to calculate kinetic energy \( KE \) is: - \( KE = \frac{1}{2} m v^2 \) - where \( m \) is the mass - \( v \) is the velocity.As an object speeds up, its kinetic energy increases because velocity is squared in the formula. In our problem, when the car reaches the bottom of the driveway, the potential energy it had at the top is transformed into kinetic energy as it moves faster. This conversion is how we determine the car鈥檚 speed at the end.
Potential Energy
Potential energy is energy stored by an object due to its position. For objects near Earth, this is often gravitational potential energy, calculated by the formula:- \( PE = m \cdot g \cdot h \) - where \( m \) is the mass, - \( g \) is the gravitational pull \( (9.8 \text{ m/s}^2) \), - and \( h \) is the height above ground.In the inclined plane scenario, the height \( h \) is found using the sine function because the driveway is inclined. The stored energy due to the car's position at the top translates into kinetic energy as it moves down.
Work-Energy Principle
The work-energy principle states that the work done on an object is equal to the change in its kinetic energy. This means if work is done on a car, it will change the car's speed because its kinetic energy changes. The formula relating work to energy is:- \( W = \Delta KE \).In the problem, when the car moves down the incline without friction, all potential energy becomes kinetic energy (without loss). When factoring in friction, the work done by friction reduces the energy available to become kinetic energy, thus affecting the final speed.
Inclined Plane
An inclined plane is a flat surface tilted at an angle, like a ramp. It's a simple machine that allows a smaller force over a longer distance to move objects to a different height. Special considerations on an incline:
  • The force in parallel to the incline affects motion.
  • Gravitational force splits into two components: parallel \( \text{and perpendicular}\) to the incline.
For our car on the driveway, only the component of gravity along the plane's direction causes motion, which is why we use \( h = 15 \cdot \sin(20^{\circ}) \) to find the effective height and potential energy.
Friction Force
Friction is the force that opposes motion between two surfaces. It converts kinetic energy to heat, reducing the object's energy available for continued motion. Frictional force \( f \) takes away from the energy that goes into increasing speed.In calculations:- The work done by friction \( W_f \) is negative, - \( W_f = - f \cdot d \), - \( f \) being the frictional force, - \( d \) being the distance traveled.In our scenario (b), friction significantly reduces the car鈥檚 speed at the bottom of the driveway, illustrating how real-world conditions can affect motion differently than ideal conditions.

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Most popular questions from this chapter

Compute the work done against gravity by a pump that discharges 600 liters of fuel oil into a tank \(20 \mathrm{~m}\) above the pump's intake. One cubic centimeter of fuel oil has a mass of \(0.82 \mathrm{~g}\). One liter is \(1000 \mathrm{~cm}^{3}\). The mass lifted is $$ (600 \text { liters })\left(1000 \frac{\mathrm{cm}^{3}}{\text { liter }}\right)\left(0.82 \frac{\mathrm{g}}{\mathrm{cm}^{3}}\right)=492000 \mathrm{~g}=492 \mathrm{~kg} $$ The lifting work is then $$ \text { Work }=(m g)(h)=\left(492 \mathrm{~kg} \times 9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(20 \mathrm{~m})=96 \mathrm{~kJ} $$

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?

A \(300-\mathrm{g}\) object slides \(80 \mathrm{~cm}\) in a straight line along a horizontal tabletop. How much work is done in overcoming friction between the object and the table if the coefficient of kinetic friction is \(0.20\) ? First find the friction force. Since the normal force equals the weight of the object, $$ F_{\mathrm{f}}=\mu_{k} F_{N}=(0.20)(0.300 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)=0.588 \mathrm{~N} $$ The work done overcoming friction is \(F_{\mathrm{t}} s \cos \theta\). Because the friction force is opposite in direction to the displacement, \(\theta=180^{\circ} .\) Therefore, $$ \text { Work }=F_{f} s \cos 180^{\circ}=(0.588 \mathrm{~N})(0.80 \mathrm{~m})(-1)=-0.47 \mathrm{~J} $$ The work is negative because the friction force is oppositely directed to the displacement; it slows the object and it decreases the object's kinetic energy, or more to the point, it opposes the motion.

A force of \(1.50 \mathrm{~N}\) acts on a \(0.20-\mathrm{kg}\) cart so as to uniformly accelerate it along a straight air track. The track and force are horizontal and in line. How fast is the cart going after acceleration from rest through \(30 \mathrm{~cm}\), if friction is negligible? The work done by the force causes, and is equal to, the increase in \(\mathrm{KE}\) of the cart. Therefore, $$ \text { Work done }=(\mathrm{KE})_{\text {end }}-(\mathrm{KE})_{\text {start }} \quad \text { or } \quad F s \cos 0^{\circ}=\frac{1}{2} m v_{f}^{2}-\mathbf{0} $$ Substituting gives $$ (1.50 \mathrm{~N})(0.30 \mathrm{~m})=\frac{1}{2}(0.20 \mathrm{~kg}) v_{f}^{2} $$ from which \(v_{f}=2.1 \mathrm{~m} / \mathrm{s}\)

How much work is done against gravity in lifting a \(3.0-\mathrm{kg}\) object through a vertical distance of \(40 \mathrm{~cm}\) ? An external force is needed to lift an object. If the object is raised at constant speed, the lifting force must equal the weight of the object. The work done by the lifting force is referred to as work done against gravity. Because the lifting force is \(m g\), where \(m\) is the mass of the object, $$ \text { Work }=(m g)(h)(\cos \theta)=(3.0 \mathrm{~kg} \times 9.81 \mathrm{~N})(0.40 \mathrm{~m})(1)=12 \mathrm{~J} $$ In general, the work done against gravity in lifting an object of mass \(m\) through a vertical distance \(h\) is \(m g h\).
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