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Chapter 8: Problem 12

A packing crute with mass \(80.0 \mathrm{~kg}\) is at rest on a horicontal. frictionlcss surface. At \(t=0\) a net horizontal force in the \(+x\) -dircction is applied to the crate. The force has a constant value of \(80.0 \mathrm{~N}\) for \(12.0 \mathrm{~s}\) and then decrcases lincarly with time so it becomes \(7 \mathrm{cro}\) after an ad. Jitional \(6.00 \mathrm{~s}\). What is the final speed of the crate. \(18.0 \mathrm{~s}\) after the force was first applied?

Short Answer

Expert verified
The final speed of the crate 18.0s after the force was first applied is \(15m/s\).

Step by step solution

01

Calculate the acceleration when force is constant

The net force acting on the crate is constant \(80.0N\) for \(12.0s\) . Using Newton's second law \(F = ma\), the acceleration \(a\) can be calculated as: \(a = F/m = 80N/80kg = 1m/s^2\)
02

Calculation of speed after 12s

The initial speed \(u\) of the crate is 0 (as it is at rest), and the acceleration \(a\) is \(1m/s^2\) for a duration \(t\) of \(12s\). Using the first equation of motion \(v = u + at\), where \(v\) is the final speed, the speed of the crate after \(12s\) can be calculated as \(v = 0 + 1m/s^2 * 12s = 12m/s\).
03

Calculate the acceleration after 12s

After \(12s\), the force decreases linearly from \(80N\) to \(0N\) over an additional \(6s\). The average force during these \(6s\) is \((80N + 0N)/2 = 40N\). So, the acceleration during these \(6s\) is \(a = F/m = 40N/80kg = 0.5m/s^2\)
04

Calculation of final speed

Using the first equation of motion again, with initial speed \(u\) as \(12m/s\), the acceleration \(a\) as \(0.5m/s^2\), and a duration \(t\) of \(6s\), the final speed of the crate can be calculated as: \(v = u + at = 12m/s + 0.5m/s^2 * 6s = 15m/s\)

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Key Concepts

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

Newton's Second Law
Newton's second law is a fundamental principle used to understand how an object will move when a force is applied. It states that the acceleration of an object is directly proportional to the net force acting upon it and inversely proportional to its mass (\( F = ma \) where F is force, m is mass, and a is acceleration). This relationship provides the means to calculate how objects will accelerate in response to various forces.

In the context of the exercise, we're given that a constant force of 80.0N acts upon an 80.0 kg crate, resulting in an acceleration of 1m/s虏. Newton's second law simplifies this scenario to a straightforward calculation where the mass of the crate serves as a direct link between the applied force and the resulting acceleration. Understanding this law is crucial because it allows us to predict the motion of the crate during the first 12 seconds and beyond as various forces are applied.
First Equation of Motion
The first equation of motion (\(v = u + at\) where v is the final velocity, u is the initial velocity, a is acceleration, and t is time) is indispensable when analyzing the motion of objects moving with uniform acceleration. It's a linear equation that connects the initial velocity, acceleration, and time with the final velocity of an object.

This equation comes into play not once, but twice in our exercise. Initially, when the crate accelerates from rest under a uniform force, and later when the force, and thus acceleration, changes. During the first 12 seconds, the crate's velocity increases linearly from 0 to 12m/s. The power of this equation lies in its simplicity, allowing students to quickly calculate changes in velocity given constant acceleration. It is also used to determine the final speed of the crate when the force鈥攁nd accordingly, the acceleration鈥攕tarts to decrease linearly.
Linearly Decreasing Force
Understanding how a linearly decreasing force affects an object's motion is critical to solving problems involving variable forces. When a force decreases linearly over time, the average force over that period can be used to find an 'average' acceleration if constant within that interval. This concept simplifies the second phase of our exercise, where the force on the crate reduces from 80N to 0N over 6 seconds.

By calculating the average force (40N in our scenario) during this period, we can derive an average acceleration and apply the first equation of motion once again to determine the final velocity of the crate after 18 seconds. It is essential to note that this only works because the force decreases at a steady rate, forming a straightforward case for using average values to find the solution.

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Most popular questions from this chapter

An \(8.00 \mathrm{~kg}\) block of wood sits at the cdge of a frictionless table, \(2.20 \mathrm{~m}\) above the floor. \(\mathrm{A} 0.500 \mathrm{~kg}\) blob of clay slides along the length of the table with a speed of \(24.0 \mathrm{~m} / \mathrm{s},\) strikes the block of wood, and sticks to it. The combined object leaves the edge of the table and travels to the floor. What horizontal distance has the combined object traveled when it reaches the floor?

Pluto and Charon. Pluto's diameter is approximately \(2370 \mathrm{~km}\), and the diameter of its satellite Charon is \(1250 \mathrm{~km}\). Although the distance varies, they are often about \(19,700 \mathrm{~km}\) apart, center to center. Assuming that both Pluto and Charon have the same composition and hence the same average density, find the location of the center of mass of this system relative to the center of Pluto.

\(A 12.0 \mathrm{~kg}\) shell is launched at an angle of \(55.0^{\circ}\) above the horizontal with an initial speed of \(150 \mathrm{~m} / \mathrm{s}\). At its highest point, the shell explodes into two fragments, one three times heavier than the other. The two fragments reach the ground at the same time. Ignore air resistance. If the heavier fragment lands back at the point from which the shell was launched, where will the lighter fragment land, and how much energy was relcased in the explosion?

You have three identical, uniform. square pieces of wood. cach with sidc Icngth \(L\). You stack the three picces of wood at the cdge of the horizontal top of a table. The first block cxtends a distance \(L / 4\) past the edge of the table. The next block extends a distance \(L / 4\) past the edge of the first block, so a distance \(L / 2\) past the edge of the table. The third block cxtcnds a distance \(L / 4\) past the codgc of the block bcncath it. so \(3 L / 4\) past the edge of the table. The stack is unstable if the center of mass of the stack extends beyond the edge of the table. Calculate the horizontal location of the center of mass of the three-block stack.

A Variable-Mass Raindrop. In a rocket-propulsion problem the mass is variable. Another such problem is a raindrop falling through a cloud of small water droplets. Some of these small droplets adhere to the raindrop, thereby increasing its mass as it falls. The force on the raindrop is $$ F_{\mathrm{ext}}=\frac{d p}{d t}=m \frac{d v}{d t}+v \frac{d m}{d t} $$ Suppose the mass of the raindrop depends on the distance \(x\) that it has fallen. Then \(m=k x,\) where \(k\) is a constant, and \(d m / d t=k u .\) This gives, since \(F_{\text {cut }}=m g_{*}\) $$ m g=m \frac{d v}{d t}+v(k v) $$ Or, dividing by \(k_{*}\) $$ x_{B}=x \frac{d v}{d t}+v^{2} $$ This is a differential equation that has a solution of the form \(v=a t\) where \(a\) is the acceleration and is constant. Take the initial velocity of the raindrop to be zero. (a) Using the proposed solution for \(v\). find the acccleration \(a\). (b) Find the distance the raindrop has fallen in \(t=3.00 \mathrm{~s} .(\mathrm{c})\) Given that \(k=2.00 \mathrm{~g} / \mathrm{m},\) find the mass of the raindrop at \(t=3.00 \mathrm{~s}\). (For many more intriguing aspects of this problem, see K. S. Krane, American Jourmal of Physics, Vol. 49(1981) . pp. \(113-117 .\) )
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