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Chapter 6: Problem 77

One end of a horizontal spring with force constant 130.0 N/m is attached to a vertical wall. A 4.00-kg block sitting on the floor is placed against the spring. The coefficient of kinetic friction between the block and the floor is \(\mu_k = 0.400\). You apply a constant force \(\overrightarrow{F}\) to the block. \(\overrightarrow{F}\) has magnitude \(F = 82.0\) N and is directed toward the wall. At the instant that the spring is compressed 80.0 cm, what are (a) the speed of the block, and (b) the magnitude and direction of the block's acceleration?

Short Answer

Expert verified
(a) The speed of the block is approximately 2.39 m/s. (b) The acceleration is 9.42 m/s虏 towards the wall.

Step by step solution

01

Identify the Forces Involved

First, identify all forces acting on the block: the applied force \( F \), the spring force \( F_s \), and the kinetic friction force \( f_k \). The spring force is given by \( F_s = kx \), where \( k = 130.0 \) N/m and \( x = 0.80 \) m. Calculate \( F_s \).
02

Calculate the Spring Force

The spring force \( F_s \) is calculated as follows:\[ F_s = kx = 130.0 \, \text{N/m} \times 0.80 \, \text{m} = 104 \, \text{N} \]
03

Determine the Frictional Force

Calculate the frictional force \( f_k \), using \( f_k = \mu_k N \). The normal force \( N \) is equal to the gravitational force \( mg \). Therefore:\[ f_k = \mu_k mg = 0.400 \times 4.00 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 15.68 \, \text{N} \]
04

Apply Newton's Second Law

Using Newton's second law, determine the net force on the block:The net force \( F_{\text{net}} \) is given by the equation:\[ F_{\text{net}} = F - F_s - f_k \]Substitute the given and calculated values:\[ F_{\text{net}} = 82.0 \, \text{N} - 104 \, \text{N} - 15.68 \, \text{N} = -37.68 \, \text{N} \]This shows the net force is in the direction opposite to \( F \).
05

Calculate the Acceleration

Use \( F_{\text{net}} = ma \) to find the acceleration \( a \) of the block.Rearranging gives:\[ a = \frac{F_{\text{net}}}{m} = \frac{-37.68 \, \text{N}}{4.00 \, \text{kg}} = -9.42 \, \text{m/s}^2 \]Thus, the block is accelerating towards the wall with magnitude 9.42 m/s虏.
06

Determine the Speed Using Kinetic Energy

Assume the compression started from rest and involves work-energy theorem:The work done on the block is the difference in kinetic and potential energy:\[ \frac{1}{2} m v^2 = F \times x - \frac{1}{2} k x^2 - f_k \times x \]Simplifying and solving for \( v \):\[ \frac{1}{2} \times 4.00 \times v^2 = 82.0 \times 0.8 - \frac{1}{2} \times 130.0 \times 0.8^2 - 15.68 \times 0.8 \]\[ 2.00 \times v^2 = 65.6 - 41.6 - 12.544 \]\[ 2.00 \times v^2 = 11.456 \]\[ v = \sqrt{5.728} \approx 2.39 \, \text{m/s} \]

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Key Concepts

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

Kinetic Friction
Kinetic friction acts as a force resisting motion between a moving object and a surface. In this exercise, the frictional force is present because the block is moving against the floor. The coefficient of kinetic friction, represented by \( \mu_k \), is a measure of how much friction the two surfaces have. A high \( \mu_k \) means more friction, while a low \( \mu_k \) means less. In this case, \( \mu_k = 0.400 \), indicating that the block experiences moderate resistance as it moves.
To calculate the frictional force \( f_k \), you need to know the normal force \( N \) acting on the object. Here, \( N \) equals the gravitational force because the block is on a flat surface, providing no vertical component to change it. Hence, we use the formula:
\[ f_k = \mu_k N = \mu_k mg \]
The calculated frictional force in this problem was 15.68 N, opposing the motion of the block. This force needs to be overcome by any applied force to keep the block moving.
Spring Force
The spring force is a restoring force exerted by a spring due to its deformation. This force follows Hooke's law, given by \( F_s = kx \), where \( k \) is the spring constant and \( x \) is the displacement from its equilibrium position. The spring constant \( k = 130.0 \ \text{N/m} \) indicates how stiff the spring is; a higher \( k \) means a stiffer spring.
In the exercise, the spring is compressed by 0.80 m. Thus, the spring exerts a force calculated as:
\[ F_s = 130.0 \ \text{N/m} \times 0.80 \ \text{m} = 104 \ \text{N} \]
This spring force acts to push the block back towards its original position. It challenges the applied force and contributes to the net force calculation that determines the block's acceleration and motion. When considering the block's movement, we need to account for this spring force acting against any applied forces.
Work-Energy Theorem
The work-energy theorem connects work done by forces to changes in kinetic energy. It states that the work done on an object equals the change in its kinetic energy. The formula encapsulates this relationship:
\[ W = \Delta KE = \frac{1}{2} m v^2 - \frac{1}{2} m v_0^2 \]
where \( v_0 \) is the initial speed.
For the block in this problem, starting from rest (i.e., \( v_0 = 0 \)), we evaluate how much energy remains after taking into account the work done by various forces, including the applied force, spring force, and frictional force. The equation is adjusted as:
\[ \frac{1}{2} m v^2 = F \times x - \frac{1}{2} k x^2 - f_k \times x \]
By solving this equation, we determined that the block's speed after compressing the spring 80 cm is approximately 2.39 m/s. This outcome shows how energy from the applied force was spent overcoming spring force and kinetic friction to set the block into motion.

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Most popular questions from this chapter

Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 1.80 \(\times\) 10\(^6\) N, one 14\(^\circ\) west of north and the other 14\(^\circ\) east of north, as they pull the tanker 0.75 km toward the north. What is the total work they do on the supertanker?

While doing a chin-up, a man lifts his body 0.40 m. (a) How much work must the man do per kilogram of body mass? (b) The muscles involved in doing a chin-up can generate about 70 J of work per kilogram of muscle mass. If the man can just barely do a 0.40-m chin-up, what percentage of his body's mass do these muscles constitute? (For comparison, the \(total\) percentage of muscle in a typical 70-kg man with 14% body fat is about 43%.) (c) Repeat part (b) for the man's young son, who has arms half as long as his father's but whose muscles can also generate 70 J of work per kilogram of muscle mass. (d) Adults and children have about the same percentage of muscle in their bodies. Explain why children can commonly do chin-ups more easily than their fathers.

A 5.00-kg block is moving at \(\upsilon_0\) \(=\) 6.00 m/s along a frictionless, horizontal surface toward a spring with force constant \(k\) = 500 N/m that is attached to a wall (\(\textbf{Fig. P6.79}\)). The spring has negligible mass. (a) Find the maximum distance the spring will be compressed. (b) If the spring is to compress by no more than 0.150 m, what should be the maximum value of \(\upsilon_0\)?

Consider a spring that does not obey Hooke's law very faithfully. One end of the spring is fixed. To keep the spring stretched or compressed an amount \(x\), a force along the \(x\)-axis with \(x\)-component \(F_x = kx - bx^2 + cx^3\) must be applied to the free end. Here \(k = 100 \, \mathrm {N/m}\), \(b = 700 \, \mathrm {N/m{^2}}\), and \(c = 12,000 \, \mathrm{N/m}^3\). Note that \(x > 0\) when the spring is stretched and \(x< 0\) when it is compressed. How much work must be done (a) to stretch this spring by 0.050 m from its unstretched length? (b) To \(compress\) this spring by 0.050 m from its unstretched length? (c) Is it easier to stretch or compress this spring? Explain why in terms of the dependence of \(F_x\) on \(x\). (Many real springs behave qualitatively in the same way.)

You are asked to design spring bumpers for the walls of a parking garage. A freely rolling 1200-kg car moving at 0.65 m/s is to compress the spring no more than 0.090 m before stopping. What should be the force constant of the spring? Assume that the spring has negligible mass.
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