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(I) How much work must be done to stop a 1300 -kg car traveling at 95\(\mathrm { km } / \mathrm { h }\) ?

Short Answer

Expert verified
The work required to stop the car is 452,645.45 Joules.

Step by step solution

01

Convert Velocity to Meters per Second

The car's speed is given in kilometers per hour (km/h). We need to convert this to meters per second (m/s) for our calculations. Use the conversion factor: \(1 \text{ km/h} = \frac{1}{3.6} \text{ m/s}\).Thus, \(95 \text{ km/h} = 95 \times \frac{1}{3.6} \approx 26.39 \text{ m/s}\).
02

Determine Initial Kinetic Energy

The work done to stop the car is equal to the car's initial kinetic energy, which is given by the formula:\[ KE = \frac{1}{2} m v^2 \]where \(m\) is the mass of the car and \(v\) is its velocity.Substitute the known values:\[ KE = \frac{1}{2} \times 1300 \times (26.39)^2 \approx 452645.45 \text{ Joules} \]
03

Calculate Work Done to Stop the Car

The work done to stop the car is equal to the negative of its initial kinetic energy (since the car is brought to rest):\[ W = - KE = -452645.45 \text{ Joules} \]
04

Consider the Sign of Work Done

Since the work done is calculated to bring the car to a stop, it is done against the car's initial motion and thus should be a negative value. However, the magnitude of work required is typically reported as a positive value:\( W = 452645.45 \text{ Joules} \)

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Key Concepts

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

Work-Energy Principle
The work-energy principle is a fundamental concept in physics. It says that the work done on an object is equal to the change in its kinetic energy. This principle is crucial when analyzing situations where forces cause motion changes, such as stopping a car. Changing an object’s velocity requires work, either adding energy to make it move faster or removing energy to slow it down. In this exercise, the work done to stop the car equals its initial kinetic energy before coming to rest. Since the car stops, the final kinetic energy is zero, and the work done is thus equal to the negative of its initial kinetic energy. This is important because it aligns with the fact that energy doesn’t disappear but transforms from kinetic energy to other forms, like heat due to friction. The equation used is:
  • Work done \[ W = - KE_{initial} \]
  • The negative sign indicates work is done against the car’s motion.
Remember, a negative work value simply indicates the direction relative to the force applied. Here, the magnitude of work indicates the total energy required to stop the car. Overall, the work-energy principle helps understand energy transfer during motion.
Velocity Conversion
Velocity conversion is essential in physics as it ensures calculations are performed in consistent units. For this exercise, we convert the car's speed from kilometers per hour (km/h) to meters per second (m/s). The standard conversion factor is:
  • 1 km/h = \[ \frac{1}{3.6} \] m/s
This conversion is crucial because many physics formulas, particularly those involving kinetic energy, require measurements in the metric system's standard units, including m/s for velocity. To convert 95 km/h to m/s:
  • Multiply 95 by \( \frac{1}{3.6} \) to get \( 26.39 \) m/s.
This provides an accurate speed in m/s, ensuring calculations like kinetic energy or work done are correct. Misalignment of units can lead to significant computational errors. Always check units when dealing with physics problems to ensure all aspects of the calculation are consistent.
Kilograms to Joules Conversion
Kilograms to Joules conversion is not direct because they measure different things. Kilograms (kg) represent mass, while Joules (J) represent energy. In physics, we often relate mass to energy using the kinetic energy formula,
  • \[ KE = \frac{1}{2} mv^2 \]
  • \(m\) is in kilograms, and \(v\) is in meters per second.
To find kinetic energy in Joules:
  • Use the mass of the object and its velocity to calculate kinetic energy.
  • Calculate with: \( KE = \frac{1}{2} \times 1300 \times (26.39)^2 = 452645.45 \) Joules
This equation shows how mass and velocity interact to produce energy, expressed in Joules. It’s important to do this conversion correctly, as it directly impacts the energy-related calculations required to solve the problem. Understanding the relationship between mass, velocity, and energy crucially links different units into practical applications such as determining the work needed to stop a moving object.

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Most popular questions from this chapter

(II) The net force along the linear path of a particle of mass 480 g has been measured at 10.0 -cm intervals, starting at \(x = 0.0 ,\) to be \(26.0,28.5,28.8,29.6,32.8,40.1,46.6,42.2 ,\) \(48.8,52.6,55.8,60.2,60.6,58.2,53.7,50.3,45.6,45.2,43.2\) \(38.9,35.1,30.8,27.2,21.0,22.2 ,\) and \(18.6 ,\) all in newtons. Determine the total work done on the particle over this entire range.

We usually neglect the mass of a spring if it is small compared to the mass attached to it. But in some applications, the mass of the spring must be taken into account. Consider a spring of unstretched length \(\ell\) and mass \(M_{\mathrm{S}}\) uniformly distributed along the length of the spring. A mass \(m\) is attached to the end of the spring. One end of the spring is fixed and the mass \(m\) is allowed to vibrate horizontally without friction (Fig. \(7-30\) ). Each point on the spring moves with a velocity proportional to the distance from that point to the fixed end. For example, if the mass on the end moves with speed \(v_{0}\), the midpoint of the spring moves with speed \(v_{0} / 2 .\) Show that the kinetic energy of the mass plus spring when the mass is moving with velocity \(v\) is $$ K=\frac{1}{2} M v^{2} $$ where \(M=m+\frac{1}{3} M_{\mathrm{S}}\) is the "effective mass" of the system. [Hint: Let \(D\) be the total length of the stretched spring. Then the velocity of a mass \(d m\) of a spring of length \(d x\) located at \(x\) is \(v(x)=v_{0}(x / D) .\) Note also that $$ d m=d x\left(M_{\mathrm{S}} / D\right) $$.

What is the dot product of \(\overrightarrow{\mathbf{A}}=2.0 x^{2} \hat{\mathbf{i}}-4.0 x \hat{\mathbf{j}}+5.0 \hat{\mathbf{k}}\) and \(\overrightarrow{\mathbf{B}}=11.0 \hat{\mathbf{i}}+2.5 x \hat{\mathbf{j}}\) ?

An elevator cable breaks when a \(925-\mathrm{kg}\) elevator is \(22.5 \mathrm{~m}\) above the top of a huge spring \((k=\) \(\left.8.00 \times 10^{4} \mathrm{~N} / \mathrm{m}\right)\) at the bottom of the shaft. Calculate (a) the work done by gravity on the elevator before it hits the spring; (b) the speed of the elevator just before striking the spring; \((c)\) the amount the spring compresses (note that here work is done by both the spring and gravity).

A cyclist starts from rest and coasts down a \(4.0^{\circ}\) hill. The mass of the cyclist plus bicycle is \(85 \mathrm{~kg} .\) After the cyclist has traveled \(250 \mathrm{~m},\) (a) what was the net work done by gravity on the cyclist? (b) How fast is the cyclist going? Ignore air resistance.

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