/*! 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 34 A 1500-kg vehicle traveling at \... [FREE SOLUTION] | 91Ó°ÊÓ

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A 1500-kg vehicle traveling at \(60 \mathrm{kph}\) collides head-on with a \(1000-\mathrm{kg}\) vehicle traveling at \(90 \mathrm{kph}\). If they come to rest immediately after impact, determine the increase in internal energy, taking both vehicles as the system.

Short Answer

Expert verified
The increase in internal energy is 521,305.44 J.

Step by step solution

01

Convert velocities to meters per second

Convert the given velocities from kilometers per hour (kph) to meters per second (m/s). The conversion is done using the factor \(1 \mathrm{km/h} = \frac{5}{18} \mathrm{m/s}\).
02

Calculate initial velocities

For the 1500-kg vehicle: \(60 \mathrm{kph} = 60 \times \frac{5}{18} = 16.67 \mathrm{m/s}\). For the 1000-kg vehicle: \(90 \mathrm{kph} = 90 \times \frac{5}{18} = 25.00 \mathrm{m/s}\).
03

Determine initial kinetic energy of each vehicle

Use the kinetic energy formula \[ E_k = \frac{1}{2}mv^2 \]. For the 1500-kg vehicle: \[ E_{k1} = \frac{1}{2} \times 1500 \times (16.67)^2 \approx 208,805.44 \mathrm{J} \]. For the 1000-kg vehicle: \[ E_{k2} = \frac{1}{2} \times 1000 \times (25.00)^2 \approx 312,500 \mathrm{J} \].
04

Calculate total initial kinetic energy

Add the initial kinetic energies of both vehicles: \[ E_{k_total} = E_{k1} + E_{k2} = 208,805.44 + 312,500 = 521,305.44 \mathrm{J} \].
05

Determine final kinetic energy

Since the vehicles come to rest immediately after impact, their final kinetic energy is zero: \[ E_{k_{final}} = 0 \mathrm{J} \].
06

Calculate the increase in internal energy

The increase in internal energy is equal to the loss of kinetic energy. \[ \Delta E_{internal} = E_{k_total} - E_{k_{final}} = 521,305.44 - 0 = 521,305.44 \mathrm{J} \].

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Key Concepts

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

kinetic energy
Before understanding the collision, it's crucial to know what kinetic energy is. Kinetic energy is the energy an object has due to its motion. The formula to calculate kinetic energy is \( E_k = \frac{1}{2} m v^2 \). This means the kinetic energy depends on both the mass (m) and the velocity (v) of the object.
When calculating kinetic energy:
  • Double the velocity, and the kinetic energy increases four times.
  • Double the mass, and the kinetic energy doubles too.
It’s essential to convert all units into standard units – typically kilograms for mass and meters per second for velocity. In our exercise, we converted the speeds from kilometers per hour (kph) to meters per second (m/s). By correctly converting and applying the formula, we derived the initial kinetic energies of both the vehicles before the collision.
collision energy transformation
A collision results in various transformations of energy. During a head-on collision, the kinetic energies of the colliding vehicles interact. When two cars collide and come to rest immediately after impact, their kinetic energy gets converted into other forms of energy.
  • Heat due to the friction between the car parts.
  • Sound waves produced during the impact.
  • Deformation energy – that is, the energy used to deform the vehicles.
In our exercise, both cars come to rest upon collision. This means their final kinetic energy is zero \( E_{k_{final}} = 0 \). Thus, the kinetic energies calculated initially become internal energy upon impact. This sort of energy transformation is common in perfectly inelastic collisions where colliding objects stick together or come to rest immediately.
conservation of energy
The principle of conservation of energy states that energy cannot be created or destroyed; it can only change forms.
During the collision of the two vehicles, the total initial kinetic energy was transformed into internal energy, showcasing the conservation of energy.
To find the increase in internal energy, we simply calculated the total initial kinetic energy and subtracted the final kinetic energy, which was zero. This transformed energy accounted for all the factors after the collision like heat, sound, and deformation.
Here’s a step by step summary:
  • Calculate the initial kinetic energy of both vehicles.
  • Note that the final kinetic energy is zero since the cars come to a stop.
  • The increase in internal energy equals the initial total kinetic energy.
The conservation of energy helps us understand how energy changes from one form to another, ensuring that the total energy remains constant within a system.

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Most popular questions from this chapter

Complete the following table if \(g=9.81 \mathrm{~m} / \mathrm{s}^{2}\) and \(V=10 \mathrm{~m}^{3}\). $$ \begin{array}{|l|c|c|c|c|c|} { } {}{|}{} & v\left(\mathrm{~m}^{3} / \mathrm{kg}\right) & \rho\left(\mathrm{kg} / \mathrm{m}^{3}\right) & \gamma\left(\mathrm{N} / \mathrm{m}^{3}\right) & m(\mathrm{~kg}) & W(\mathrm{~N}) \\ \hline(a) & 20 & & & & \\ (b) & & 2 & & & \\ (c) & & & 4 & & \\ (d) & & & & 100 & \\ (e) & & & & & 100 \\ \hline \end{array} $$

Calculate the density, specific weight, mass, and weight of a body that occupies \(200 \mathrm{ft}^{3}\) if its specific volume is \(10 \mathrm{ft}^{3} / \mathrm{lbm}\). The quantities will not be calculated in the order asked for. The mass is $$ m=\frac{V}{v}=\frac{200 \mathrm{ft}^{3}}{10 \mathrm{ft}^{3} / \mathrm{lbm}}=20 \mathrm{lbm} $$ The density is $$ \rho=\frac{1}{v}=\frac{1}{10}=0.1 \mathrm{lbm} / \mathrm{ft}^{3} $$ The weight is, using \(g=32.2 \mathrm{ft} / \mathrm{s}^{2}\), $$ W=\frac{m}{g_{c}} g=\frac{20 \mathrm{lbm}}{32.2 \mathrm{lbm}-\mathrm{ft} / \mathrm{s}^{2}-\mathrm{lbf}}\left(32.2 \frac{\mathrm{ft}}{\mathrm{s}^{2}}\right)=20 \mathrm{lbf} $$ Finally, the specific weight is calculated to be $$ w=\frac{W}{V}=\frac{20}{200}=0.1 \mathrm{lbf} / \mathrm{ft}^{3} $$ Note that using English units, (1.6) would be written as $$ \gamma=\frac{\rho}{g_{c}} g=\left(\frac{0.1 \mathrm{lbm} / \mathrm{ft}^{3}}{32.2 \mathrm{lbm}-\mathrm{ft} / \mathrm{s}^{2}-\mathrm{lbf}}\right)\left(32.2 \mathrm{ft} / \mathrm{s}^{2}\right)=0.1 \mathrm{lbf} / \mathrm{ft}^{3} $$

Determine the deceleration of \((a)\) a \(2200-\mathrm{kg}\) car and \((b)\) a \(1100-\mathrm{kg}\) car, if the brakes are suddenly applied so that all four tires slide on a horizontal surface. The coefficient of friction \(\eta=0.6\) on the dry asphalt. ( \(\eta=F / N\) where \(N\) is the normal force and \(F\) is the frictional force.)

Determine the weight of a mass of \(10 \mathrm{~kg}\) at a location where the acceleration of gravity is \(9.77 \mathrm{~m} / \mathrm{s}^{2}\).

A tube can be inserted into the top of a pipe transporting liquids, providing the pressure is relatively low, so that the liquid fills the tube at height \(h\). Determine the pressure in a water pipe if the water seeks a level at height \(h=6 \mathrm{ft}\) above the center of the pipe. The pressure is found from (1.11) to be $$ P=\gamma h=(62.4)(6)=374 \mathrm{lbf} / \mathrm{ft}^{2} \quad \text { or } \quad 2.60 \text { psi gage } $$

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