# A 2.50-kg mass is pushed against a horizontal spring of force constant 23.0 N/cm on a...

## Question:

A 2.50-kg mass is pushed against a horizontal spring of force constant 23.0 N/cm on a frictionless air table. The spring is attached to the tabletop, and the mass is not attached to the spring in any way. When the spring has been compressed enough to store 13.5 J of potential energy in it, the mass is suddenly released from rest.

(a) Find the greatest speed the mass reaches.

(b) When does this occur?

(c) What is the greatest acceleration of the mass

(d) When does it occur?

## Kinetic energy:

The term kinetic energy can be defined as the energy that is occupied by the body when it is in motion. Its measurable unit is in the Joules. The formula for the kinetic energy is given by {eq}k = \dfrac{1}{2}m{v^2} {/eq}

## Answer and Explanation:

**Given data**:

- The mass is {eq}m = 2.50\,{\rm{kg}} {/eq}

- The force constant is {eq}K = 23.0\,{\rm{N/cm}} = 2300\,{\rm{m}} {/eq}

- The potential energy stored in the spring is {eq}{U_1} = 13.5\,{\rm{J}} {/eq}

**(a)**

The expression for the work energy theorem is given by

{eq}{k_1} + {U_1} = {k_2} + {U_2} {/eq}

- Here {eq}k = \dfrac{1}{2}m{v^2} {/eq} is the kinetic energy.

- Here {eq}U = \dfrac{1}{2}k{x^2} {/eq}is the potential energy.

The expression for the conservation of energy is given by

{eq}{k_2} = {U_1} {/eq}

Substituting the values in the above equation as,

{eq}\begin{align*} {k_2} &= {U_1}\\ {k_2} &= 13.5\,{\rm{J}} \end{align*} {/eq}

Substituting the values in the kinetic energy equation

{eq}\begin{align*} k &= \dfrac{1}{2}m{v^2}\\ v &= \sqrt {\dfrac{{2 \times k}}{m}} \\ v &= \sqrt {\dfrac{{2 \times 13.5}}{{2.5}}} \\ v &= 2.078\,{\rm{m/s}} \end{align*} {/eq}

Thus the greatest speed of the mass reaches is {eq}v = 2.078\,{\rm{m/s}} {/eq}

**(b)**

This occurs only when the kinetic energy is maximum. According to the conservation of energy, the mass is starting to move its potential energy is converted into kinetic energy.

**(c)**

The expression for the potential energy is given by

{eq}\begin{align*} U &= \dfrac{1}{2}K{x^2}\\ x &= \sqrt {\dfrac{{2 \times U}}{K}} \end{align*} {/eq}

Substituting the values in the above equation as,

{eq}\begin{align*} x &= \sqrt {\dfrac{{2 \times U}}{K}} \\ x &= \sqrt {\dfrac{{2 \times 13.5}}{{2300}}} \\ x &= 2.25 \times {10^{ - 3}}\,{\rm{m}} \end{align*} {/eq}

The expression for the force stretched by the spring is given by

{eq}F = K \times x {/eq}

The expression for the acceleration of the mass is given by

{eq}a = \dfrac{F}{m} {/eq}

Substituting the values in the above equation as,

{eq}\begin{align*} a &= \dfrac{F}{m}\\ a &= \dfrac{{k \times x}}{m}\\ a &= \dfrac{{2300 \times \left( {2.25 \times {{10}^{ - 3}}} \right)}}{{2.5}}\\ a &= 20.7\,{\rm{m/}}{{\rm{s}}^2} \end{align*} {/eq}

Thus the acceleration of the mass is {eq}a = 20.7\,{\rm{m/}}{{\rm{s}}^2} {/eq}

**(d)**

This occurs only when the force acting on the mass by the spring is greatest.

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from Geography 101: Human & Cultural Geography

Chapter 13 / Lesson 9