Question

A block of mass m is projected straight upward by a strong spring whose stiffness is${\mathbf{k}}_{\mathbf{s}}$ . When the block is a height ${\mathit{y}}_{\mathbf{1}}$above the floor, it is travelling upward at speed${\mathbf{v}}_{\mathbf{1}}$ , and the spring is compressed an amount ${\mathbf{s}}_{\mathbf{1}}$. A short time later the block is at height ${\mathit{y}}_{\mathbf{2}}$, travelling upward at speed ${\mathbf{v}}_{\mathbf{2}}$, and the spring is compressed an amount ${\mathbf{s}}_{\mathbf{2}}$. Assume that thermal transfer of energy (microscopic work) Q between the block and the air is negligible. For each of the following choices of system, write the Energy Principle in the update form${\mathbf{E}}_{\mathbf{f}}\mathbf{=}{\mathbf{E}}_{\mathbf{i}}\mathbf{+}\mathbf{W}$ . (a) The block, the spring and the Earth; (b) the block plus spring; (c) the block alone.

Short answer

$$\begin{gathered} \left(\mathrm{a}\right)\frac{1}{2}{\mathrm{mv}}_1^2+\frac{1}{2}{\mathrm{k}}_{\mathrm{s}}{\mathrm{s}}_1^2-{\mathrm{mgy}}_1=\frac{1}{2}{\mathrm{mv}}_2^2+\frac{1}{2}{\mathrm{k}}_{\mathrm{s}}{\mathrm{s}}_2^2-{\mathrm{mgy}}_2 \\ \left(\mathrm{b}\right)\frac{1}{2}{\mathrm{mv}}_1^2+\frac{1}{2}{\mathrm{k}}_{\mathrm{s}}{\mathrm{s}}_1^2-\mathrm{mg}({\mathrm{y}}_1-{\mathrm{y}}_2)=\frac{1}{2}{\mathrm{mv}}_2^2+\frac{1}{2}{\mathrm{k}}_{\mathrm{s}}{\mathrm{s}}_2^2 \\ \left(\mathrm{c}\right)\frac{1}{2}{\mathrm{mv}}_1^2+\frac{1}{2}{\mathrm{k}}_{\mathrm{s}}{\mathrm{s}}_1^2-\mathrm{mg}({\mathrm{y}}_1-{\mathrm{y}}_2)-\frac{1}{2}{\mathrm{k}}_{\mathrm{s}}({\mathrm{s}}_2^2-{\mathrm{s}}_1^2)=\frac{1}{2}{\mathrm{mv}}_2^2 \end{gathered}$$

Step-by-step solution

Identification of the given data

The given data can be listed below as,

• The mass of the block is, m.
• The stiffness of spring is,${\mathrm{k}}_{\mathrm{s}}$ .
• The initial height of the block is,$y_1$ .
• The initial speed of the block is,$v_1$ .
• The initial compression of the spring is,$s_1$ .
• The final height of the block is, $y_2$.
• The final speed of the block is, $v_2$.
• The final compression of the spring is, $s_2$.

Explanation of the conservation of energy in the case of the spring-mass system, Spring energy, and Kinetic energy

In the case of a spring-mass system, the mechanical energy is conserved if the applied force is constant. So, the initial energy of the spring-mass system is the same as that of the final energy of the spring-mass system. It can be expressed as follows,

${\mathbf{E}}_{\mathbf{i}}\mathbf{=}{\mathbf{E}}_{\mathbf{f}}$…… (1)

Here${\mathrm{E}}_{\mathrm{j}}$ is the initial energy of the system and${\mathrm{E}}_{\mathrm{f}}$ is the final energy of the system.

The expression for the spring energy is as follows,

$\mathrm{PE}=\frac{1}{2}{\mathrm{kx}}^2$

Here,k is the spring stiffness and x is the compression or extension length.

The expression for the kinetic energy is as follows,

$\mathrm{KE}=\frac{1}{2}{\mathrm{mv}}^2$

Here, m is the mass of the object and v is the velocity with which the object is moving.

Determination of the Energy Principle for the block, the spring, and the Earth

(a)

It is known that the block, the spring, and the Earth are taken as the system, and hence there is no action of any external force on them

the expression for the energy by using equation (1) is,

${\mathrm{KE}}_{\mathrm{i}}+{\mathrm{SE}}_{\mathrm{i}}+{\mathrm{PE}}_{\mathrm{i}}={\mathrm{KE}}_{\mathrm{f}}+{\mathrm{SE}}_{\mathrm{f}}+{\mathrm{PE}}_{\mathrm{f}}$

Here, left side of the equation shows the total initial energy of the system, which includes initial kinetic energy (role="math" localid="1657793548953" ${\mathrm{KE}}_{\mathrm{j}}$), initial spring energy (role="math" localid="1657793568370" ${\mathrm{SE}}_{\mathrm{j}}$), initial potential energy (role="math" localid="1657793588172" ${\mathrm{PE}}_{\mathrm{j}}$), final kinetic energy (${\mathrm{KE}}_{\mathrm{j}}$), final spring energy (${\mathrm{SE}}_{\mathrm{j}}$), and final potential energy (${\mathrm{PE}}_{\mathrm{j}}$).

Substituting all the values in the above expression,

$\frac{1}{2}{\mathrm{mv}}_1^2+\frac{1}{2}{\mathrm{k}}_{\mathrm{s}}{\mathrm{s}}_1^2-{\mathrm{mgy}}_1=\frac{1}{2}{\mathrm{mv}}_2^2+\frac{1}{2}{\mathrm{k}}_{\mathrm{s}}{\mathrm{s}}_2^2-{\mathrm{mgy}}_2$

Here, g is the acceleration due to gravity.

Thus, the Energy Principle for the block, the spring, and the Earth,$\frac{1}{2}{\mathrm{mv}}_1^2+\frac{1}{2}{\mathrm{k}}_{\mathrm{s}}{\mathrm{s}}_1^2-{\mathrm{mgy}}_1=\frac{1}{2}{\mathrm{mv}}_2^2+\frac{1}{2}{\mathrm{k}}_{\mathrm{s}}{\mathrm{s}}_2^2-{\mathrm{mgy}}_2.$

Determination of the Energy Principle for the block and the spring

(b)

It is known that the block and the spring are taken as the system, and hence there is one external force that is acting on the system, and that is due to gravity. So, some work has been done by this force.

Write the expression for the energy by using equation (1).

${\mathrm{KE}}_{\mathrm{i}}+{\mathrm{SE}}_{\mathrm{i}}+\mathrm{W}={\mathrm{KE}}_{\mathrm{f}}+{\mathrm{SE}}_{\mathrm{f}}$

Here W is the work done by gravity which is equal to $\mathrm{mg}({\mathrm{y}}_2-{\mathrm{y}}_1)$.

Substitute all the values in the above expression,

$\frac{1}{2}{\mathrm{mv}}_1^2+\frac{1}{2}{\mathrm{k}}_{\mathrm{s}}{\mathrm{s}}_1^2-\mathrm{mg}({\mathrm{y}}_1-{\mathrm{y}}_2)=\frac{1}{2}{\mathrm{mv}}_2^2+\frac{1}{2}{\mathrm{k}}_{\mathrm{s}}{\mathrm{s}}_2^2$

Thus, the Energy Principle for the block, and the spring is,$\frac{1}{2}{\mathrm{mv}}_1^2+\frac{1}{2}{\mathrm{k}}_{\mathrm{s}}{\mathrm{s}}_1^2-\mathrm{mg}({\mathrm{y}}_1-{\mathrm{y}}_2)=\frac{1}{2}{\mathrm{mv}}_2^2+\frac{1}{2}{\mathrm{k}}_{\mathrm{s}}{\mathrm{s}}_2^2.$

Determination of the Energy Principle for the block only

(c)

It is known that the block is taken as the system, and hence there is two external force that is acting on the system and which are spring force and force due to gravity. So, some work has been done by these forces.

Write the expression for the energy by using equation (1).

${\mathrm{KE}}_{\mathrm{i}}+{\mathrm{SE}}_{\mathrm{i}}+\mathrm{W}+{\mathrm{W}}_{\mathrm{s}}={\mathrm{KE}}_{\mathrm{f}}+{\mathrm{SE}}_{\mathrm{f}}$

Here, W is the work done by the gravity, which is equal to $mg(y_2-y_1)$and$W_s$ is the work done by the spring, which is equal to $\frac{1}{2}{k}_s(s_2^2-s_1^2).$

Substitute all the values in the above expression,

$\frac{1}{2}{\mathrm{mv}}_1^2+1/2{\mathrm{k}}_{\mathrm{s}}{\mathrm{s}}_1^2-\mathrm{mg}({\mathrm{y}}_1-{\mathrm{y}}_2)-\frac{1}{2}{\mathrm{k}}_{\mathrm{s}}({\mathrm{s}}_2^2-{\mathrm{s}}_1^2)=\frac{1}{2}{\mathrm{mv}}_2^2$

Thus, the Energy Principle for the block only is,$\frac{1}{2}{\mathrm{mv}}_1^2+\frac{1}{2}{\mathrm{k}}_{\mathrm{s}}{\mathrm{s}}_1^2-\mathrm{mg}({\mathrm{y}}_1-{\mathrm{y}}_2)-\frac{1}{2}{\mathrm{k}}_{\mathrm{s}}({\mathrm{s}}_2^2-{\mathrm{s}}_1^2)=\frac{1}{2}{\mathrm{mv}}_2^2.$