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Chapter 6: Problem 36

A 2.0 -kg box and a 3.0 -kg box on a perfectly smooth horizontal floor have a spring of force constant 250 \(\mathrm{N} / \mathrm{m}\) compressed between them. If the initial compression of the spring is \(6.0 \mathrm{cm},\) find the acceleration of each box the instant after they are released. Be sure to include free-body diagrams of each box as part of your solution.

Short Answer

Expert verified
The 2 kg box accelerates at \(7.5 \text{ m/s}^2\) and the 3 kg box at \(5 \text{ m/s}^2\).

Step by step solution

01

Identify Given Values

We have two boxes on a smooth horizontal floor. The boxes have masses of \(2.0 \text{ kg}\) and \(3.0 \text{ kg}\). The spring between them has a force constant of \(k = 250 \text{ N/m}\) and is initially compressed by \(6.0 \text{ cm} = 0.06 \text{ m}\). We need to find the accelerations of each box right after they are released.
02

Use Hooke's Law to Find Force

According to Hooke's Law, the force exerted by a spring is given by \( F = kx \), where \( k \) is the spring constant and \( x \) is the displacement. Substitute the values: \( F = 250 \text{ N/m} \times 0.06 \text{ m} = 15 \text{ N} \). This force is exerted equally on both boxes in opposite directions.
03

Apply Newton's Second Law to Each Box

For the 2.0 kg box, apply Newton's second law: \( F = ma \). The force acting on it is 15 N, so \( 15 \text{ N} = 2.0 \text{ kg} \times a_1 \). Solving for \( a_1 \), we have \( a_1 = \frac{15 \text{ N}}{2.0 \text{ kg}} = 7.5 \text{ m/s}^2 \).For the 3.0 kg box, similarly, \( F = ma \). The force is the same (15 N but in the opposite direction), so \( 15 \text{ N} = 3.0 \text{ kg} \times a_2 \). Solving for \( a_2 \), we have \( a_2 = \frac{15 \text{ N}}{3.0 \text{ kg}} = 5 \text{ m/s}^2 \).
04

Draw Free-Body Diagrams

For each box, draw a free-body diagram. For the 2 kg box, draw a force arrow pointing to the right labeled \(15 \text{ N}\). For the 3 kg box, draw a force arrow pointing to the left also labeled \(15 \text{ N}\). These represent the forces exerted by the spring immediately after release.

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Key Concepts

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

Hooke's Law
Hooke's Law is a principle of physics that states the force required to compress or extend a spring is proportional to the distance the spring is stretched or compressed.
This relationship is expressed in the formula:
  • \( F = kx \)
Here, \( F \) is the force exerted by the spring, \( k \) is the spring's constant (measured in Newtons per meter), and \( x \) is the displacement from the spring's original length.
If you have a spring constant of \( 250 \text{ N/m} \) and compress the spring by \( 0.06 \text{ m} \), the spring exerts a force of \( 15 \text{ N} \).
In the context of our problem, this force is what propels the boxes apart once released.
spring force
The spring force is the push or pull exerted by a spring.
It acts in the opposite direction of the displacement that caused it. In our case, with two boxes compressed by a spring, each box experiences a force in the opposite direction to the compression.
The spring, with a force constant of \( 250 \text{ N/m} \), applies a force of \( 15 \text{ N} \) to each box upon release.
  • The spring force on the 2 kg box pushes it to the right.
  • The spring force on the 3 kg box pushes it to the left.
Understanding this directional force is crucial for determining the motion of each box.
acceleration calculations
Acceleration is a measure of how quickly an object changes its velocity. According to Newton鈥檚 Second Law, acceleration\( (a) \) can be calculated using the formula:
  • \( a = \frac{F}{m} \)
where \( F \) is the force applied and \( m \) is the mass of the object.
In our exercise:
  • The 2.0 kg box experiences an acceleration of \( 7.5 \text{ m/s}^2 \) due to a 15 N force.
  • The 3.0 kg box experiences an acceleration of \( 5 \text{ m/s}^2 \) with the same force, but applied in the opposite direction.
This difference in acceleration occurs because Newton's Second Law considers both force and mass. With smaller mass on the 2 kg box, it accelerates more.
free-body diagrams
Free-body diagrams are essential tools used to represent the forces acting on an object.
They help visualize the magnitude and direction of all forces, aiding in understanding the motion that results.
  • For the 2 kg box, the free-body diagram shows a single arrow pointing to the right, labeled 15 N.
  • For the 3 kg box, the arrow points to the left with the same 15 N label.
These diagrams are a simplified way to model the physical scenario, focusing on the forces at play at a snapshot in time, right when the spring is released.
Using these diagrams aids in applying Newton's laws effectively to solve problems related to forces and motion.

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Most popular questions from this chapter

CALC A Spring with Mass. We usually ignore the kinetic energy of the moving coils of a spring, but let's try to get a reasonable approximation to this. Consider a spring of mass \(M,\) equilibrium length \(L_{0},\) and spring constant \(k .\) The work done to stretch or compress the spring by a distance \(L\) is \(\frac{1}{2} k X^{2}\) , where \(X=L-L-L_{0}\) . Consider a spring, as described above, that has one end fixed and the other end moving with speed \(v .\) Assume that the speed of points along the length of the spring varies linearly with distance \(l\) from the fixed end. Assume also that the mass \(M\) of the spring is distributed uniformly along the length of the spring. (a) Calculate the kinetic energy of the spring in terms of the \(M\) and \(v .\) (Hint: Divide the spring into pieces of length \(d l ;\) find the speed of each pivide in terms of \(l, v,\) and \(L ;\) find the mass of each piece in terms of \(d l, M,\) and \(L ;\) and integrate from 0 to \(L .\) The result is \(n o t \frac{1}{2} M v^{2},\) since not all of the spring moves with the same speed.) In a spring gun, a spring of mass 0.243 \(\mathrm{kg}\) and force constant 3200 \(\mathrm{N} / \mathrm{m}\) is compressed 2.50 \(\mathrm{cm}\) from its unstretched length. When the trigger is pulled, the spring pushes horizon- tally on a 0.053 -kg ball. The work done by friction is negligible. Calculate the ball's speed when the spring reaches its uncom- pressed length (b) ignoring the mass of the spring and (c) includ- ing, using the results of part (a), the mass of the spring. (d) In part (c), what is the final kinetic energy of the ball and of the spring?

A loaded grocery cart is rolling across a parking lot in a strong wind. You apply a constant force \(\vec{F}=(30 \mathrm{N}) \hat{\imath}-(40 \mathrm{N}) \hat{\mathrm{J}}\) to the cart as it undergoes a displacement \(\vec{s}=(-9.0 \mathrm{m}) \hat{\boldsymbol{\imath}}-(3.0 \mathrm{m}) \hat{\boldsymbol{J}}\) . How much work does the force you apply do on the grocery cart?

Meteor Crater. About \(50,000\) years ago, a meteor crashed into the earth near present-day Flagstaff, Arizona. Measurements from 2005 estimate that this meteor had a mass of about \(1.4 \times 10^{8}\) kg (around \(150,000\) tons) and hit the ground at a speed of 12 \(\mathrm{km} / \mathrm{s}\) . (a) How much kinetic energy did this meteor deliver to the ground? (b) How does this energy compare to the energy released by a \(1.0-\) megaton nuclear bomb? (A megaton bomb releases the same amount of energy as a million tons of TNT, and 1.0 ton of TNT releases \(4.184 \times 10^{9}\) J of energy.)

A little red wagon with mass 7.00 \(\mathrm{kg}\) moves in a straight line on a frictionless horizontal surface. It has an initial speed of 4.00 \(\mathrm{m} / \mathrm{s}\) and then is pushed 3.0 \(\mathrm{m}\) in the direction of the initial velocity by a force with a magnitude of 10.0 \(\mathrm{N}\) . (a) Use the work-energy theorem to calculare the wagon's final speed. (b) Cal- culate the acceleration produced by the force. Use this acceleration in the kinematic relationships of Chapter 2 to calculate the wagon's final speed. Compare this result to that calculated in part (a).

A block of ice with mass 2.00 \(\mathrm{kg}\) slides 0.750 \(\mathrm{m}\) down an inclined plane that slopes downward at an angle of \(36.9^{\circ}\) below the horizontal. If the block of ice starts from rest, what is its final speed? You can ignore friction.
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