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Chapter 10: Problem 57

Block and Horizontal Spring , a \(2.5 \mathrm{~kg}\) block slides head on into a spring with a spring constant of \(320 \mathrm{~N} / \mathrm{m}\). When the block stops, it has compressed the spring by \(7.5 \mathrm{~cm}\). The coefficient of kinetic friction between the block and the horizontal surface is \(0.25 .\) While the block is in contact with the spring and being brought to rest, what are (a) the work done by the spring force and (b) the increase in thermal energy of the block-floor system? (c) What is the block's speed just as the block reaches the spring?

Short Answer

Expert verified
The work done by the spring force is 0.9 J. The increase in thermal energy of the block-floor system is 0.46 J. The block's speed just as it reaches the spring is 0.305 m/s.

Step by step solution

01

- Convert Units

Convert the compression distance of the spring from cm to meters: 7.5 cm = 0.075 m
02

- Increase in Thermal Energy

The increase in thermal energy of the block-floor system is equal to the work done by the kinetic friction force: The kinetic friction force Welcro is also the work done by the kinetic friction force and the work done by the block applied by the friction force as a proportion of the overall friction force that the box applies to it. Show this many times when pennywheels come around.The expression f = FN * Uk= 0.23 J as per formula.
03

- Calculate Speed Just Before Compression

Use the work-energy principle: The initial kinetic energy of the block is equal to the work done by the spring plus the work done by friction. Initial Kinetic Energy = Potential Energy in the Spring + Thermal Energy Increase where v is the speed of the block right before it starts compressing the spring. Rearrange the formula and solve for v.

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Key Concepts

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

Work-Energy Principle
Understanding the work-energy principle is key to solving this problem. This principle states that the work done on an object is equal to its change in kinetic energy. This principle can be written as \[ W_{total} = \triangle KE \]. In our scenario, the block's kinetic energy transforms into potential energy stored in the spring and thermal energy due to friction. Formulas used will relate kinetic energy \[ KE = \frac{1}{2} mv^2 \] and work done by forces, either compressing the spring or overcoming friction.
Kinetic Friction
Kinetic friction occurs when two objects slide against each other. The force due to kinetic friction \[ f_k = \text{渭}_k \times N \] where \text{渭}_k is the coefficient of kinetic friction (0.25 in this case) and N is the normal force. In our case, the normal force equals the gravitational force (\text{mg}). Therefore, \[ f_k = 0.25 \times 2.5 \times 9.8 = 6.125 \]. This force does work in opposing the block's motion, converting kinetic energy into thermal energy.
Spring Constant
The spring constant (\text{k}) relates the force exerted by a spring to its compression or extension. In this problem, \text{k} is given as 320 \text{N/m}. The force exerted by the spring is \[ F = -kx \] which means the work done by the spring when compressed can be calculated using \[ W_{spring} = \frac{1}{2} k x^2 \]. We plug in x = 0.075 m (converted from 7.5 cm), to find the energy stored: \[ W_{spring} = \frac{1}{2} \times 320 \times (0.075)^2 = 0.9 \text{J} \].
Potential Energy
Potential energy in a spring (Elastic Potential Energy) is given by \[ PE_{spring} = \frac{1}{2} k x^2 \]. This formula indicates how the energy is stored when the spring is compressed or stretched. In our problem, the potential energy stored in the spring at maximum compression would be equal to the work done to compress it, calculated as 0.9 J. This stored energy will be used to bring the block to rest.
Thermal Energy
Thermal energy arises in this problem from kinetic friction between the block and floor. The work done by friction force gets converted into thermal energy, calculated by \[ W_{friction} = f_k \times d \]. If we consider the block moves until it stops compressing the spring, the distance it travels under friction force before full compression needs to be estimated. This in turn helps us estimate the thermal energy increase in the block-floor system.
Kinetic Energy
Initially, the block has kinetic energy which is used up in compressing the spring and overcoming friction. The initial kinetic energy is given by \[ KE = \frac{1}{2} mv^2 \]. To find the speed (v) at the point where the block just touches the spring, use the energy balance: \[ KE_{initial} = PE_{spring} + W_{friction} \]. Rewriting it in terms of v: \[ \frac{1}{2} mv^2 = 0.9 J + W_{friction} \]. By solving for v, we get the speed at the instant the block starts compressing the spring, illustrating how kinetic energy is distributed into other forms.

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Most popular questions from this chapter

Two Snowy Peaks Two snowy peaks are \(850 \mathrm{~m}\) and \(750 \mathrm{~m}\) above the valley between them. A ski run extends down from the top of the higher peak and then back up to the top of the lower one, with a total length of \(3.2 \mathrm{~km}\) and an average slope of \(30^{\circ}\) (Fig. 10-49). (a) A skier starts from rest at the top of the higher peak. At what speed will he arrive at the top of the lower peak if he coasts without using ski poles? Ignore friction. (b) Approximately what coefficient of kinetic friction between snow and skis would make him stop just at the top of the lower peak?

Elastic Collision of Cart A cart with mass \(340 \mathrm{~g}\) moving on a frictionless linear air track at an initial speed of \(1.2 \mathrm{~m} / \mathrm{s}\) undergoes an elastic collision with an initially stationary cart of unknown mass. After the collision, the first cart continues in its original direction at \(0.66 \mathrm{~m} / \mathrm{s}\). (a) What is the mass of the second cart? (b) What is its speed after impact? (c) What is the speed of the two-cart center of mass?

Chain on Table a chain is held on a frictionless table with one-fourth of its length hanging over the edge. If the chain has length \(L\) and mass \(m\), how much work is required to pull the hanging part back onto the table?

Two Blocks and a Spring A block of mass \(m_{A}=2.0 \mathrm{~kg}\) slides along a frictionless table with a speed of \(10 \mathrm{~m} / \mathrm{s}\). Directly in front of it, and moving in the same direction, is a block of mass \(m_{B}=5.0 \mathrm{~kg}\) moving at \(3.0 \mathrm{~m} / \mathrm{s}\). A massless spring with spring constant \(k=1120\) \(\mathrm{N} / \mathrm{m}\) is attached to the near side of \(m_{B}\), as shown in Fig. \(10-57\). When the blocks collide, what is the maximum compression of the spring? (Hint: At the moment of maximum compression of the spring, the two blocks move as one. Find the velocity by noting that the collision is completely inelastic at this point.)

. Frisbee A \(75 \mathrm{~g}\) Frisbee is thrown from a point \(1.1 \mathrm{~m}\) above the ground with a speed of \(12 \mathrm{~m} / \mathrm{s}\). When it has reached a height of \(2.1 \mathrm{~m}\), its speed is \(10.5 \mathrm{~m} / \mathrm{s}\). What was the reduction in the mechanical energy of the Frisbee- Earth system because of air drag?
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