## Textbooks & Solution Manuals

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

## Tip our Team

Our Website is free to use.
To help us grow, you can support our team with a Small Tip.

## Holooly Tables

All the data tables that you may search for.

## Holooly Help Desk

Need Help? We got you covered.

## Holooly Arabia

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

Products

## Textbooks & Solution Manuals

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

## Holooly Arabia

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

## Holooly Help Desk

Need Help? We got you covered.

## Q. 7.7

A mass m is dropped from a height h onto a linear spring of constant stiffness k and negligible mass. Determine the maximum deflection δ of the spring, and compare this value with the static spring deflection  $\delta_S$  produced by m, See Fig. 7.7. Assume that m maintains contact with the spring in its motion following the impact.

## Verified Solution

Since the velocity of m is zero at both its initial and terminal states at 1 and 3 in Fig. 7.7, the change in its kinetic energy on the path  $\mathscr{C}$ is zero. Because the mass of the spring is negligible , its kinetic energy may be ignored . Moreover, all the forces that act on m are constant, workless, or vary only with the particle’s position on $\mathscr{C}$. Therefore, the work–energy principle may be applied .

The total work done by the forces acting on m from its initial position 1 to its final position 3 on  $\mathscr{C}$  is determined by (7.21) in which  $\mathscr{C}=\mathscr{C}_1\cup \mathscr{C}_2.$  We note that the instantaneous impulsive force of the spring does no work on m. The free body diagrams in Fig. 7.7 show that the force acting on m over  $\mathscr{C}_1$  is  $F_1$ = mgi and over $\mathscr{C}_2$  is  $F_2=(mg – kx_2)$  i. Hence,

$\mathscr{W}=\int_{\mathscr{C}_1}^{}{F.dx} + \int_{\mathscr{C_2}}^{}{F.dx}=\int_{0}^{h}{mgdx_1} + \int_{0}^{\delta}{(mg – kx_2)dx_2}.$            (7.42a)

The work–energy equation (7.36) thus yields

$\mathscr{W}=mgh + mg \delta – \frac{1}{2}k {\delta}^{2} = \Delta {K}=0,$                 (7.42b)

which determines the following deflection of the spring:

$\delta=\frac{mg}{k} + \sqrt{(\frac{mg}{k})^{2} + \frac{2mgh}{k}}.$             (7.42c)

The static deflection that would result from the weight alone is  ${\delta}_s=mg/k.$  Use of this relation in (7.42c) gives

$\delta={\delta}_S + \sqrt{{\delta}^{2}_S + 2{\delta}_Sh} \geq 2 {\delta}_S.$                     (7.42d)

This formula shows that the dynamic deflection δ is not less than twice the static deflection  ${\delta}_S$.  In particular, if m is released just at the top of the spring so that h = 0, then  $\delta = 2{\delta}_S.$

The foregoing solution has illustrated the application of the work–energy equation when the work is calculated by use of the path integrals in (7.42a). Alternatively, we recognize that the work  $\mathscr{W}_g$  done by the gravitational force acting over the entire path from 1 to 3 in the direction of the displacement is  given by   $\mathscr{W}_g=mg(h + \delta)$;  the work  $\mathscr{W}_e$  done by the elastic spring force acting on m over the path from 2 to 3 is $\mathscr{W}_e=\frac{1}{2}k{\delta}^{2}$;  and the impulsive force imposed on m at state 2 is workless. Hence, the total work done on m is   $\mathscr{W}=\mathscr{W}_g + \mathscr{W}_e=mg(h + \delta) – \frac{1}{2}k {\delta}^{2}$,  which is the same as (7.42b) derived above.