Question

In the block–spring arrangement of Fig.${}^{\mathbf{7}\mathbf{-}\mathbf{10}}$, the block’s mass is ${}^{\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{kg}}$and the spring constant is${}^{\mathbf{500}\mathbf{N}\mathbf{/}\mathbf{m}}$. The block is released from position${}^{{\mathbf{x}}_{\mathbf{i}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{300}\mathbf{}\mathbf{m}}$. What is (a) the block’s speed at${}^{\mathbf{x}\mathbf{=}\mathbf{0}}$, (b) the work done by the spring when the block reaches${}^{\mathbf{x}\mathbf{=}\mathbf{0}}$, (c) the instantaneous power due to the spring at the release point ${}^{{\mathbf{x}}_{\mathbf{i}}}$, (d) the instantaneous power at${}^{\mathbf{x}\mathbf{=}\mathbf{0}}$, and (e) the block’s position when the power is maximum?

Short answer

a. The block’s speed at ${}^{\mathrm{x}=0}$is${}^{\mathrm{v}=3.35\mathrm{m}/\mathrm{s}}$.

b. The work done by the spring when the block reaches ${}^{x=0}$is ${}^{\mathrm{w}=22.5\mathrm{J}}$.

c. Instantaneous power due to the spring at the release pointlocalid="1654233250413" ${}^{{\mathrm{x}}_{\mathrm{i}}}$is${}^{\mathrm{P}=0\mathrm{w}}$.

d. The instantaneous power at x = 0 is${}^{\mathrm{P}=0\mathrm{w}}$.

e. At ${}^{\mathrm{x}=0.212,}$the power is maximum.

Step-by-step solution

Given information

It is given that,

1. Mass of the block is 4.00kg.

2. Spring constant is 500N/m.

3. Distance${}^{{\mathrm{x}}_{\mathrm{i}}=0.300\mathrm{m}}$.

Determining the concept

The problem is based on the Hooke’s law. It states that the displacement or size of a deformation is directly proportional to the deforming force or load for relatively modest deformations of an object. Here the concept of spring’s potential energy and kinetic energy of the object can be applied. spring’s potential energy is the stored energy in a compressible or stretchable objects such as spring, rubber. Using the energy conservation concept, the speed of the block can be found. From the potential energy of the spring, the work can be measured. Further, using the equation of power, which relates the force and velocity, the power at the given positions can be computed.

Required formulae are as follow:

The force by Hooke’s law can be written as

$\mathrm{F}=-\mathrm{kx}$ (I)

The potential energy is given by,

$\mathrm{U}=\frac{1{\mathrm{kx}}^2}{2}$ (ii)

Kinetic energy is given by,

$$\begin{gathered} \mathrm{k}=\frac{1}{2}{\mathrm{mv}}^2 \\ \mathrm{p}=\mathrm{Fv} \end{gathered}$$

Where, F is the force, P is the power, v is the velocity, m is the mass, x is the displacement and k is the spring constant.

(a) determining the block’s speed at x=0

Use the conservation of energy equation (iii) to find the speed,

$$\begin{gathered} \frac{1}{2}{\mathrm{mv}}^2=\frac{1}{2}\mathrm{k}({{\mathrm{x}}_1}^2-{{\mathrm{x}}_1}^2) \\ \mathrm{When},{\mathrm{x}}_{\mathrm{f}}=0, \\ {\mathrm{v}}^2=\frac{{{\mathrm{kx}}_{\mathrm{i}}}^2}{\mathrm{m}} \\ \mathrm{v}=\sqrt{\frac{{{\mathrm{kx}}_{\mathrm{i}}}^2}{\mathrm{m}}}=\sqrt{\frac{\left(500\right){(0.300)}^2}{4.00}} \\ \mathrm{v}=3.35\mathrm{m}/\mathrm{s} \\ \mathrm{Hence},\mathrm{the}\mathrm{block}\text{'}\mathrm{s}\mathrm{speed}\mathrm{at}\mathrm{x}=0\mathrm{is}\mathrm{v}=3.35\mathrm{m}/\mathrm{s} \end{gathered}$$

(b) Determining the work done by the spring when the block reachesx=0

As work done by the spring is equal to its potential energy. Thus equation (ii) can be written as,

$\mathrm{W}=\frac{1}{2}{{\mathrm{kx}}_{\mathrm{i}}}^2$

$$\begin{gathered} \mathrm{W}=\frac{1}{2}\mathrm{k}{(0.300)}^2=\frac{1}{2}\left(500\right){(0.300)}^2\mathrm{W}=22.5\mathrm{J} \\ \mathrm{Hence},\mathrm{the}\mathrm{work}\mathrm{done}\mathrm{by}\mathrm{the}\mathrm{spring}\mathrm{when}\mathrm{the}\mathrm{block}\mathrm{reaches}\mathrm{x}=0\mathrm{is}\mathrm{w}=22.5\mathrm{J} \end{gathered}$$

(c) Determining the instantaneous power due to the spring at the release pointxi

Power is force times velocity.

So, using equation (iv)

$$\begin{gathered} \mathrm{p}=\mathrm{Fv} \\ \mathrm{p}=\left({\mathrm{kx}}_{\mathrm{i}}\right){\mathrm{v}}_{\mathrm{i}} \\ =\left(500\right)(0.300)\left(0\right) \\ =0\mathrm{w} \\ \mathrm{p}=0\mathrm{w} \\ \mathrm{Hence},\mathrm{the}\mathrm{instantaneous}\mathrm{power}\mathrm{due}\mathrm{to}\mathrm{the}\mathrm{spring}\mathrm{at}\mathrm{the}\mathrm{release}\mathrm{point} \\ {\mathrm{x}}_{\mathrm{i}}\mathrm{is}\mathrm{p}=0\mathrm{w}. \end{gathered}$$

(d) determining the instantaneous power atx=0

As calculated above, write that instantaneous power at x is zero ${}^{\mathrm{P}=\mathrm{FV}}$

At${}^{\mathrm{x}=0}$, the force on the particle is equal to 0. So, the instantaneous power would be equal to zero.

Hence, the instantaneous power at x = 0 islocalid="1654235103909" ${}^{\mathrm{P}=0\mathrm{W}}$.

(e) Determining the block’s position when the power is maximum

Now, find the maximum power by differentiating P with respect to x and equating it to zero,

$$\begin{gathered} \frac{\mathrm{dP}}{\mathrm{dx}}=\frac{{\mathrm{k}}^2({{\mathrm{x}}_{\mathrm{i}}}^2-2{\mathrm{x}}^2)}{\sqrt{{\displaystyle \frac{\mathrm{k}({{\mathrm{x}}_{\mathrm{i}}}^2-{\mathrm{x}}^2)}{\mathrm{m}}}}} \\ =0 \\ {\mathrm{k}}^2({{\mathrm{x}}_{\mathrm{i}}}^2-2{\mathrm{x}}^2)=0 \\ ({{\mathrm{x}}_{\mathrm{i}}}^2-2{\mathrm{x}}^2)=0 \\ \mathrm{x}=\frac{{\mathrm{x}}_{\mathrm{i}}}{\sqrt{2}} \\ \mathrm{x}=0.212\mathrm{m} \\ \mathrm{so},\mathrm{the}\mathrm{power}\mathrm{is}\mathrm{maximum}\mathrm{at}\mathrm{x}=0.212\mathrm{m}. \end{gathered}$$