Question

A $\mathbf{0}.\mathbf{250}-\mathbf{kg}$block resting on a frictionless, horizontal surface is attached to a spring whose force constant is $\mathbf{83}\mathbf{.}\mathbf{8}\mathbf{ }\mathbf{N}\mathbf{\text{/m}}$as in Figure P15.31. A horizontal force $\overrightarrow{\mathit{F}}$causes the spring to stretch a distance of $\mathbf{5}.\mathbf{46} \mathbf{cm}$from its equilibrium position. (a) Find the magnitude ofrole="math" localid="1660112519769" $\overrightarrow{\mathit{F}}$. (b) What is the total energy stored in the system when the spring is stretched? (c) Find the magnitude of the acceleration of the block just after the applied force is removed. (d) Find the speed of the block when it first reaches the equilibrium position. (e) If the surface is not frictionless but the block still reaches the equilibrium position, would your answer to part (d) be larger or smaller? (f) What other information would you need to know to find the actual answer to part (d) in this case? (g) What is the largest value of the coefficient of friction that would allow the block to reach the equilibrium position?

Short answer

(a) $4.575 \text{N}$

(b) $0.125 \text{J}$

(c)$18.3 {\text{m/s}}^{\text{2}}$

(d)$1.00 \text{m/s}$

(e) Smaller

(f) Coefficient of kinetic friction

(g)$0.933$

Step-by-step solution

Identification of given data

Mass of block is$m=0.250 \text{kg}$

Force constant of spring is$k=83.8 \mathrm{N}\text{/m}$

Stretch of spring isrole="math" localid="1660111082711" $x=5.46 \mathrm{cm} \text{or} \text{0.0546} \text{m}$

Significance of hook’s law

According to the physics principle known as Hooke's Law, the amount of force required to extend or compress a spring proportionally depends on the length of the spring.

$\overrightarrow{F}=-kx$ ...(i)

Where, $k$is force constant and $x$is stretch of the spring

Here, negative sign represent the restoring force always in opposite direction to the displacement of the block

(a) Determining the magnitude of F→

Only need to find the magnitude so that equation (i) will be expressed as

$F=kx$

Substituting all the values in above equation

$\begin{array}{rcl}F& =& \left(83.8 \text{N/m}\right)\times \left(0.0546 \text{m}\right)\\ & =& 4.575 \text{N}\\ & & \end{array}$

Hence the magnitude$\overrightarrow{F}$ of is$4.575 \text{N}$

(b) Determining the total energy stored in the system when the spring is stretched

Total energy stored in the system when the spring is stretched is equal to the elastic potential energy of the spring block system and it is given as follow

$E=\frac{1}{2}k{x}^2$

Substitute all the values in above equation

$\begin{array}{rcl}E& =& \frac{1}{2}\times \left(83.8 \text{N/m}\right)\times {\left(0.0546 \text{m}\right)}^2\\ & =& 0.125 \text{J}\end{array}$

Hence thetotal energy stored in the system when the spring is stretched is$0.125 \text{J}$

(c) Determining the magnitude of the acceleration of the block just after the applied force is removed

From the Newton’s second law of motion

$F=ma$

Where, $F$is applied force, $m$is mass of block and $a$is acceleration of the block

Substitute all the values in above equation

$\begin{array}{rcl}4.575 \text{N}& =& \left(0.250 \text{kg}\right)\times a\\ a& =& 18.3 {\text{m/s}}^{\text{2}}\\ & & \end{array}$

Hence the magnitude of the acceleration of the block just after the applied force is removed is$18.3 {\text{m/s}}^{\text{2}}$

(d) Determining the speed of the block when it first reaches the equilibrium position

In a basic harmonic motion, a block's equilibrium position has its maximum velocity. The block has its highest kinetic energy in the equilibrium position. The block's kinetic energy and total energy are equal in its equilibrium position.

$E=\frac{1}{2}m{v}^2$

Rearrange the above equation

$v=\sqrt{\frac{2E}{m}}$

Substitute all the values in above equation

$\begin{array}{rcl}v& =& \sqrt{\frac{2\times 0.125 \text{J}}{0.250 \text{kg}}}\\ & =& 1.00 \text{m/s}\end{array}$

Hence the speed of the block when it first reaches the equilibrium position is$1.00 \text{m/s}$

(e) If the surface is not frictionless then determining whether the answer to part (d) be larger or smaller

If the surface is not frictionless then the total potential energy of the system is not completely converting into kinetic energy. From part (d) it is showing that velocity is directly proportional to total potential energy and when the friction occurs, energy will be reduced and hence velocity will be reduced

Hence the speed of the block when it reaches the equilibrium is smaller as compared to the answer in part (d)

(f) Determining the other information needed to get the answer of part(d) when the surface is not frictionless

The friction between the block and the surface results in energy loss if the surface is frictional. To determine the block's speed, one has to know the coefficient of kinetic friction.

(g) Determining the largest value of the coefficient of friction

The expression of friction force for the block is,

$f=\mu mg$

Where, $\mu$is coefficient of friction for block and surface, $m$is mass for block and $a$is acceleration due to gravity

By using the principle of energy conservation, the work required to push the block up against the friction is equal to the potential energy stored in the spring when it is extended.

$\begin{array}{rcl}\left(f\right)\left(x\right)& =& \frac{1}{2}k{x}^2\\ \left(\mu mg\right)\left(x\right)& =& \frac{\text{1}}{\text{2}}k{x}^2\\ \mu & =& \frac{kx}{2mg}\end{array}$

Substituting the values in above equation

localid="1660112167467" $\begin{array}{rcl}\mu & =& \frac{\left(83.8 \text{N/m}\right)\times \left(0.0546 \text{m}\right)}{2\times 0.250 \text{kg}\times 9.8 {\text{m/s}}^2}\\ & =& 0.933\end{array}$

Hence the largest value of the coefficient of friction is$0.933$