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Chapter 4: Problem 11

What average force is needed to accelerate a 9.20-gram pellet from rest to \(125 \mathrm{~m} / \mathrm{s}\) over a distance of \(0.800 \mathrm{~m}\) along the barrel of a rifle?

Short Answer

Expert verified
The average force needed is 89.44 N.

Step by step solution

01

Convert Mass to Kilograms

First, convert the mass of the pellet from grams to kilograms since the standard SI unit for mass is kilograms. Given that the mass is 9.20 grams, we convert it by dividing by 1000. Thus, the mass \( m \) is \( \frac{9.20}{1000} = 0.0092 \) kg.
02

Identify Known Values and Identify the Formula

We know the final speed \( v = 125 \) m/s, the initial speed \( u = 0 \) m/s, the distance \( s = 0.800 \) m, and the mass \( m = 0.0092 \) kg. The force can be found by first determining acceleration using the equation \( v^2 = u^2 + 2as \), and then using \( F = ma \).
03

Calculate Acceleration

Rearranging the kinematic equation \( v^2 = u^2 + 2as \) to solve for acceleration \( a \), we have:\[a = \frac{v^2 - u^2}{2s} = \frac{125^2 - 0^2}{2 \cdot 0.800}\]Calculating gives:\[a = \frac{15625}{1.6} = 9765.625 \text{ m/s}^2\]
04

Calculate Force

Using Newton's second law \( F = ma \), substitute the mass \( m = 0.0092 \) kg and acceleration \( a = 9765.625 \) m/s虏 to find the force:\[F = 0.0092 \times 9765.625 = 89.44 \text{ N}\]
05

Finalize the Answer

The average force needed is calculated to be approximately 89.44 Newtons. This result is obtained by multiplying the calculated acceleration by the mass of the pellet.

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Key Concepts

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

Kinematic Equations
Kinematic equations are fundamental in physics for describing the motion of objects. They relate variables such as displacement, initial velocity, final velocity, acceleration, and time, allowing us to solve motion-related problems.
Kinematic equations assume constant acceleration. The equation used in this exercise, \( v^2 = u^2 + 2as \), is particularly useful when we don't know the time\( t \) and want to find either acceleration \( a \) or the distance \( s \).
  • \( v \) is the final velocity.
  • \( u \) is the initial velocity.
  • \( a \) is the acceleration.
  • \( s \) is the displacement.
By rearranging the equation, you can solve for the unknown variable. In this case, we solved for acceleration \( a \) using the known values of final and initial velocities and the distance traveled.
Unit Conversion in Physics
Unit conversions are crucial in physics to ensure calculations are performed accurately using SI units. This exercise involves converting mass from grams to kilograms, an essential step before using Newton's Second Law to calculate force.
Here's how you do it:
  • The mass was given in grams.
  • Since 1 kilogram = 1000 grams, divide the mass in grams by 1000 to convert to kilograms.
For instance, 9.20 grams becomes 0.0092 kilograms. Consistent units ensure the calculations align with the standard scientific framework, providing accuracy in applying physics principles.
Concept of Force
Force is a central concept in physics, defined by Newton's Second Law as the product of mass and acceleration: \( F = ma \). It quantifies the interaction that changes an object's state of motion.
In this exercise, force was calculated to understand the pellete's acceleration within the rifle barrel.
  • Mass \( m \) was found by converting from grams to kilograms.
  • Acceleration \( a \) was derived using the kinematic equation \( v^2 = u^2 + 2as \).
The final force in Newtons represents how much is needed to reach the specified velocity over the given distance. This application of Newton's Second Law illustrates the relationship between mass, force, and motion, showing how these fundamental concepts interconnect in practical scenarios.

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Most popular questions from this chapter

$$ \begin{array}{l}{\text { (II) Uphill escape ramps are sometimes provided to the }} \\ {\text { side of steep downhill highways for trucks with overheated }} \\ {\text { brakes. For a simple } 11^{\circ} \text { upward ramp, what length would be }} \\ {\text { needed for a runaway truck traveling } 140 \mathrm{km} / \mathrm{h} \text { ? Note the }}\end{array} $$ $$ \begin{array}{l}{\text { large size of your calculated length. (If sand is used for the bed }} \\ {\text { of the ramp, its length can be reduced by a factor of about } 2 .}\end{array} $$

A fisherman yanks a fish vertically out of the water with an acceleration of \(2.5 \mathrm{~m} / \mathrm{s}^{2}\) using very light fishing line that has a breaking strength of \(18 \mathrm{~N}(\approx 4 \mathrm{lb})\). The fisherman unfortunately loses the fish as the line snaps. What can you say about the mass of the fish?

(II) Using focused laser light, optical tweezers can apply a force of about 10 \(\mathrm{pN}\) to a 1.0 -\mum diameter polystyrene bead, which has a density about equal to that of water: a volume of 1.0 \(\mathrm{cm}^{3}\) has a mass of about 1.0 \(\mathrm{g}\) . Estimate the bead's acceleration in \(g^{\prime} \mathrm{s}\) .

(II) A large crate of mass 1500 \(\mathrm{kg}\) starts sliding from rest along a frictionless ramp, whose length is \(\ell\) and whose incli- nation with the horizontal is \(\theta\) (a) Determine as a function of \(\theta :(\mathrm{i})\) the acceleration \(a\) of the crate as it goes downhill, (ii) the time \(t\) to reach the bottom of the incline, (iii) the final velocity \(v\) of the crate when it reaches the bottom of the ramp, and (iv) the normal force \(F_{\mathrm{N}}\) on the crate. \((b)\) Now assume \(\ell=100 \mathrm{m}\) . Use a spreadsheet to calculate and graph \(a, t, v,\) and \(F_{\mathrm{N} \text { as functions of } \theta \text { from } \theta=0^{\circ} \text { to } 90^{\circ} \text { in }}\) \(1^{\circ}\) steps. Are your results consistent with the known result for the limiting cases \(\theta=0^{\circ}\) and \(\theta=90^{\circ} ?\)

Sketch the free-body diagram of a baseball \((a)\) at the moment it is hit by the bat, and again (b) after it has left the bat and is flying toward the outfield.
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