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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

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. \(113-117 .\) )

A system consists of two particles. \(\mathrm{At} t=0\) one particle is at the origin; the other, which has a mass of \(0.50 \mathrm{~kg}\). is on the \(y\) -axis at \(y=6,0 \mathrm{~m} .\) At \(t=0\) the center of mass of the system is on the \(y\) -axis at \(y=2.4 \mathrm{~m} .\) The velocity of the center of mass is given by \(\left(0.75 \mathrm{~m} / \mathrm{s}^{3}\right) t^{2} \hat{i}\) (a) Find the total mass of the system. (b) Find the acceleration of the center of mass at any time \(t\). (c) Find the net external force acting on the system at \(t=3.0 \mathrm{~s}\)

One \(110 \mathrm{~kg}\) football lineman is running to the right at \(2.75 \mathrm{~m} / \mathrm{s}\) while another 125 kg lineman is nunning directly toward him at \(2,60 \mathrm{~m} / \mathrm{s}\). What ure (a) the magnitude and direction of the net momentum of these two athletes, and (b) their total kinctic encrgy?

A block with mass \(0.500 \mathrm{~kg}\) sits at rest on a light but not long vertical spring that has spring constant \(80.0 \mathrm{~N} / \mathrm{m}\) and one end on the floor. (a) How much clastic potential cnergy is stored in the spring when the block is sitting at rest on it? (b) A second identical block is dropped onto the first from a height of \(4.00 \mathrm{~m}\) above the first block and sticks to it. What is the maximum elastic potential energy stored in the spring during the motion of the blocks after the collision? (c) What is the maximum distance the first block moves down after the second block has landed on it?

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.
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