Question

A $\mathbf{250}\mathbf{g}$block is dropped onto a relaxed vertical spring that has a spring constant of $\mathbf{K}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{N}\mathbf{/}\mathbf{cm}$(Fig.).The block becomes attached to the spring and compresses the spring$\mathbf{12}\mathbf{}\mathbf{cm}$before momentarily stopping. While the spring is being compressed, what work is done on the block by (a) the gravitational force on it and (b) the spring force? (c) What is the speed of the block just before it hits the spring? (Assume that friction is negligible.) (d) If the speed at impact is doubled, what is the maximum compression of the spring?

Short answer

1. Work done by gravitational force is$0.29\mathrm{J}$

1. Work done by the spring is$-1.8\mathrm{J}$
2. Speed of the object before it hit the spring is$3.5\mathrm{m}/\mathrm{s}$
3. If $V_f=2V_f$, then distance of compression is$0.23\mathrm{m}$

Step-by-step solution

Given information

Mass of the block is$m=250\mathrm{kg}$

Spring constant is$k=2.5\mathrm{N}/\mathrm{cm}=250\mathrm{N}/\mathrm{m}$

Compressed distance is$d=12\mathrm{cm}=0.12\mathrm{m}$

Determining the concept

This problem deals with the work-energy theorem which states that the net work done by the forces applied on object is equal to the change in its kinetic energy. To solve this problem, use the work-energy theorem and the equation relating work done and force.

Formulae:

$$\begin{gathered} W=Fx\cos \varphi \\ W=∆K \end{gathered}$$

Where,

F is force, xis displacement,$∆K$ is change in kinetic energy and Wis the work done.

(a) Determining the work done by gravitational force ()

$$\begin{gathered} W_g=F_{g}d\cos \varphi \\ \varphi =0° \\ mg=mgd\cos 0 \\ mg=0.25\times 9.8\times 0.12 \\ mg=0.294\mathrm{J} \\ \approx 0.29\mathrm{J} \end{gathered}$$ (i)

Hence, the work done by gravitational force is $W_g=0.29\mathrm{J}$

(b) determining the work done by the spring (Ws)

Kinetic energy of the spring is

$$\begin{gathered} K=\frac{1}{2}k{d}^2 \\ {W}_s=∆K \\ =K_i-K_f \end{gathered}$$

As the initial position of the spring is mean position,

role="math" localid="1657176614801" $$\begin{gathered} W_s=-K_f \\ {W}_s=-\frac{1}{2}k{d}^2\left(\mathrm{ii}\right) \\ {W}_s=-\frac{1}{2}\times 250\times {(0.12)}^2{W}_s=-1.8\mathrm{J} \end{gathered}$$

Hence, the work done by the spring is$W_s=-1.8\mathrm{J}$

(c) Determining the speed of the object before it hit the spring ( v )

$W_{net}=∆K=K_i-K_f$

Astheinitial position of the spring is mean position,$K_i=0$,

$$\begin{gathered} W_{net}=-K_f \\ {W}_s+W_g=-\frac{1}{2}m{v}_f^2 \\ -1.8+0.294=-\frac{1}{2}\times 250\times v_f^2 \\ {v}_f^2=\sqrt{\left(\frac{2\times 1.506}{0.25}\right)} \\ {v}_f=3.471\mathrm{m}/\mathrm{s} \\ {\mathrm{v}}_{\mathrm{f}}=3.5\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the speed of the object before it hit the spring$v=3.5\mathrm{m}/\mathrm{s}$.

(d) Determining the distance of compression d if Vf=2Vf

$$\begin{gathered} W_{net}=-Kf \\ {W}_s+W_g=-\frac{1}{2}m{v}_{{}^f}^2 \\ U\sin g{v}_f=2v_f\mathrm{and}\mathrm{equation}(iand2, \\ -\frac{1}{2}k{d}^2+mgd=-\frac{1}{2}m{\left(2v_f\right)}^2\mathrm{Solving}\mathrm{and}\mathrm{rearranging}, \\ {d}^2-\frac{2mgd}{k}-\frac{m{v}_f^2}{k}=0 \\ \mathrm{This}\mathrm{is}\mathrm{the}\mathrm{quardratic}\mathrm{equation}\mathrm{with}, \\ a=1,b=-\frac{2mg}{k},c=-\frac{m{v}_f^2}{k} \\ \mathrm{As}\mathrm{the}\mathrm{negative}\mathrm{value}\mathrm{will}\mathrm{give}\mathrm{a}\mathrm{negative}\mathrm{root},\mathrm{consider}\mathrm{only}\mathrm{the}\mathrm{positive}\mathrm{factor}, \\ d=\frac{-b+\sqrt{b^2-4ac}}{2a} \\ d=\frac{-\left(-{\displaystyle \frac{2mg}{k}}\right)+\sqrt{{\left(-{\displaystyle \frac{2mg}{k}}\right)}^2-4\left(-{\displaystyle \frac{\mathit{m}{\mathit{v}}_{\mathit{f}}^{\mathit{2}}}{\mathit{k}}}\right)}}{2\mathrm{a}} \\ d=\frac{\left({\displaystyle \frac{2mg}{k}}\right)+{\displaystyle \frac{2}{k}}\sqrt{\left({\displaystyle \frac{4m^2{g}^2}{k^2}}\right)+4{\displaystyle \frac{m{v}_f^2}{k}}}}{2} \\ d=\frac{\left({\displaystyle \frac{2mg}{k}}\right)+{\displaystyle \frac{2}{k}}\sqrt{m^2{k}^2+m{v}_f^{2}k}}{2} \\ d=\frac{mg+\sqrt{m^2{g}^2+m{v}_f^{2}k}}{k} \\ d=\frac{0.25\times 9.8+\sqrt{{\left(0.25\right)}^2{(9.8)}^2+0.25{(3.471)}^{2}250}}{250} \\ \mathrm{By}\mathrm{solving}\mathrm{the}\mathrm{equation}, \\ d=0.23\mathrm{m} \end{gathered}$$

Hence, if$v^{\text{'}}=2v$, then distance of compression is $d=0.23\mathrm{m}$.