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Chapter 10: Problem 44

Bullet Hits Wall A \(30 \mathrm{~g}\) bullet, with a horizontal velocity of \(500 \mathrm{~m} / \mathrm{s}\), comes to a stop \(12 \mathrm{~cm}\) within a solid wall. (a) What is the change in its mechanical energy? (b) What is the magnitude of the average force from the wall stopping it?

Short Answer

Expert verified
(a) The change in mechanical energy is \(3750 \text{ J}\). (b) The magnitude of the average force from the wall stopping the bullet is \(31250 \text{ N}\).

Step by step solution

01

Identify given data

The mass of the bullet is given as \(30 \text{ g} = 0.03 \text{ kg}\). The initial velocity (\(v_i\)) is \(500 \text{ m/s}\). The bullet comes to a stop, so the final velocity (\(v_f\)) is \(0 \text{ m/s}\). The stopping distance (\(d\)) is \(12 \text{ cm} = 0.12 \text{ m}\).
02

Calculate change in kinetic energy

The change in mechanical energy is essentially the initial kinetic energy of the bullet, which is converted to other forms of energy (like heat and deformation energy). The formula for kinetic energy is \(KE = \frac{1}{2}mv^2\). Thus, the change in kinetic energy is:\[\text{Change in KE} = KE_i - KE_f = \frac{1}{2}mv_i^2 - \frac{1}{2}mv_f^2 = \frac{1}{2} \cdot 0.03 \cdot (500)^2 - 0 = 3750 \text{ J}\]
03

Calculate the average stopping force

Using the work-energy principle, the work done by the stopping force is equal to the change in kinetic energy. Work done \(W = F \times d\). Rearranging the formula, we have:\[F = \frac{W}{d} = \frac{3750 \text{ J}}{0.12 \text{ m}} = 31250 \text{ N}\]

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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 that an object possesses due to its motion. This energy depends on two main factors: the mass of the object and its velocity. The formula to calculate kinetic energy is: \(KE = \frac{1}{2}mv^2\), where \(m\) is the mass and \(v\) is the velocity. In the example of the bullet hitting the wall, we used this formula to calculate the kinetic energy of the bullet before it stopped. Since the bullet comes to a stop, its final velocity is zero, meaning its final kinetic energy is also zero. Hence, the change in kinetic energy is equal to its initial kinetic energy, which in our case, was 3750 Joules.
Work-Energy Principle
The work-energy principle is a powerful concept that connects the work done on an object to its change in kinetic energy. According to this principle: \(W = \text{Change in KE}\). Here, the 'work' is the energy transferred to or from an object via the force applied over a distance. For the bullet example, the work done by the wall on the bullet (to stop it) is the same as the negative of the initial kinetic energy of the bullet, which converts to other forms of energy (like heat). The formula we use is \(W = F \times d\), where \(W\) is work, \(F\) is the average force, and \(d\) is the distance over which the force acts. By rearranging this formula, we could solve for the average force.
Average Force
Force is a push or pull exerted on an object, and it can cause the object to accelerate, decelerate, or change direction. Average force is the constant force that would produce the same change in motion as the actual varying force over the given distance. In our bullet example, we determined the average force exerted by the wall to stop the bullet. Using the work-energy principle and knowing the change in kinetic energy and the stopping distance, we found: \(F = \frac{W}{d}\). Here, the work done (\(W\)) was 3750 Joules, and the distance (\(d\)) was 0.12 meters. Plugging these values into our formula, we calculated that the average force from the wall was 31250 Newtons.

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Most popular questions from this chapter

Nonconforming Spring A certain spring is found \(n o t\) to conform to Hooke's law. The force (in newtons) it exerts when stretched a distance \(x\) (in meters) is found to have magnitude \((52.8 \mathrm{~N} / \mathrm{m}) x+\left(38.4 \mathrm{~N} / \mathrm{m}^{2}\right) x^{2}\) in the direction opposing the stretch. (a) Compute the work required to stretch the spring from \(x_{1}=0.500\) \(\mathrm{m}\) to \(x_{2}=1.00 \mathrm{~m} .\) (b) With one end of the spring fixed, a particle of mass \(2.17 \mathrm{~kg}\) is attached to the other end of the spring when it is extended by an amount \(x_{2}=1.00 \mathrm{~m}\). If the particle is then released from rest, what is its speed at the instant the spring has returned to the configuration in which the extension is \(x_{1}=0.500 \mathrm{~m} ?(\mathrm{c})\) Is the force exerted by the spring conservative or nonconservative? Explain.

Runaway Truck In Fig. \(10-33\), a runaway truck with failed brakes is moving downgrade at \(130 \mathrm{~km} / \mathrm{h}\) just before the driver steers the truck up a frictionless emergency escape ramp with an in- clination of \(15^{\circ} .\) The truck's mass is \(5000 \mathrm{~kg} .\) (a) What minimum length \(L\) must the ramp have if the truck is to stop (momentarily) along it? (Assume the truck is a particle, and justify that assumption.) Does the minimum length \(L\) increase, decrease, or remain the same if (b) the truck's mass is decreased and (c) its speed is decreased?

Worker Pushes Block A worker pushed a \(27 \mathrm{~kg}\) block \(9.2 \mathrm{~m}\) along a level floor at constant speed with a force directed \(32^{\circ}\) below the horizontal. If the coefficient of kinetic friction between block and floor was \(0.20\), what were (a) the work done by the worker's force and (b) the increase in thermal energy of the block-floor system?

What is the spring constant of a spring that stores \(25 \mathrm{~J}\) of elastic potential energy when compressed by \(7.5 \mathrm{~cm}\) from its relaxed length?

Spring at the Top of an Incline a spring with spring constant \(k=170 \mathrm{~N} / \mathrm{m}\) is at the top of a \(37.0^{\circ}\) frictionless incline. The lower end of the incline is \(1.00 \mathrm{~m}\) from the end of the spring, which is at its relaxed length. A \(2.00 \mathrm{~kg}\) canister is pushed against the spring until the spring is compressed \(0.200 \mathrm{~m}\) and released from rest. (a) What is the speed of the canister at the instant the spring returns to its relaxed length (which is when the canister loses contact with the spring)? (b) What is the speed of the canister when it reaches the lower end of the incline?
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