# Box, mass m_1, at rest against a compressed spring, spring constant k and compression of \Delta...

## Question:

Box, mass {eq}m_1 {/eq}, at rest against a compressed spring, spring constant {eq}k {/eq} and compression of {eq}\Delta x {/eq}, at the top of a ramp, height {eq}h {/eq}, is released. It slides down the ramp, angle {eq}\theta {/eq}, with friction, coefficient {eq}\mu {/eq}. Half way down the ramp it bumps into and sticks to another box, mass {eq}m_2 {/eq}, and together, they slide to the bottom. If {eq}m_1 = 2.0 \ kg, m_2= 3.0 \ kg, k = 500 \ N/m, \Delta x = 0.40 \ m, h = 5.0 \ m, \mu = 0.30, \theta = 30.0 ^o, {/eq} how long does it take both boxes to reach the bottom (starting from when the first box separates from the spring)? The spring is parallel to the ramp surface and the {eq}\Delta x {/eq} is measured parallel to the ramp surface.

## Spring Force:

The spring force is the force that is stored in the spring due to the compression or extension. The value of the force depends on the amount of extension and spring constant, and its direction is always opposite to the motion of the motion.

## Answer and Explanation: 1

** Given data **

- The mass of the box is: {eq}{m_1} = 2\;{\rm{kg}} {/eq}

- The mass of another box is: {eq}{m_2} = 3\;{\rm{kg}} {/eq}

- The value of the spring constant is: {eq}k = 500\;{{\rm{N}} {\left/ {\vphantom {{\rm{N}} {\rm{m}}}} \right. } {\rm{m}}} {/eq}

- The value of the compression is: {eq}\Delta x = 0.4\;{\rm{m}} {/eq}

- The height of the ramp is: {eq}h = 5\;{\rm{m}} {/eq}

- The coefficient of friction is: {eq}\mu = 0.3 {/eq}

- The angle of the ramp is: {eq}\theta = 30^\circ {/eq}

- The initial velocity of the second block is: {eq}{u_2} = 0 {/eq}

The potential energy stored in the spring is converted in to kinetic energy of the mass. The expression for the potential energy and kinetic energy is given as:

{eq}\begin{align*} \dfrac{1}{2}k{\left( {\Delta x} \right)^2} &= \dfrac{1}{2}{m_1}u_1^2\\ {u_1} &= \left( {\Delta x} \right) \times \sqrt {\dfrac{k}{{{m_1}}}} \end{align*} {/eq}

Substitute the value in the above equation.

{eq}\begin{align*} {u_1} &= 0.4\;{\rm{m}} \times \sqrt {\dfrac{{500\;{{\rm{N}} {\left/ {\vphantom {{\rm{N}} {\rm{m}}}} \right. } {\rm{m}}}}}{{2\;{\rm{kg}}}}} \\ {u_1} &= 6.324\;{{\rm{m}} {\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. } {\rm{s}}} \end{align*} {/eq}

The expression for the normal reaction force on the first box is given as:

{eq}N = mg\cos \theta {/eq}

The expression for the friction force on the first block is given as:

{eq}\begin{align*} {f_r} &= \mu N\\ {f_r} &= \mu mg\cos \theta \end{align*} {/eq}

The expression for the acceleration for the first block is given as:

{eq}\begin{align*} ma &= mg\sin \theta - {f_r}\\ ma &= mg\sin \theta - \mu mg\cos \theta \\ a &= g\left( {\sin \theta - \mu \cos \theta } \right) \end{align*} {/eq}

Substitute the value in the above equation.

{eq}\begin{align*} a &= 9.81 \times \left( {\sin 30^\circ - 0.3 \times \cos 30^\circ } \right)\\ a &= 2.356\;{{\rm{m}} {\left/ {\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right. } {{{\rm{s}}^{\rm{2}}}}} \end{align*} {/eq}

Let us assume the inclined distance be {eq}H. {/eq}

The expression for the inclined distance is given as:

{eq}H = \dfrac{h}{{\sin \theta }} {/eq}

The first box is halfway down the ramp it bumps into and sticks to another box. The expression for half of the inclined distance is given as:

{eq}\begin{align*} L &= \dfrac{H}{2}\\ L &= \dfrac{1}{2} \times \dfrac{h}{{\sin \theta }} \end{align*} {/eq}

Substitute the value in the above equation.

{eq}\begin{align*} L &= \dfrac{1}{2} \times \dfrac{5}{{\sin 30^\circ }}\\ L &= 5\;{\rm{m}} \end{align*} {/eq}

The expression for the velocity at the half way down the ramp is given as:

{eq}{v^2} = u_1^2 + 2aL {/eq}

Substitute the value in the above expression.

{eq}\begin{align*} v_1^2 &= {\left( {6.324\;{{\rm{m}} {\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. } {\rm{s}}}} \right)^2} + 2 \times \left( {2.356\;{{\rm{m}} {\left/ {\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right. } {{{\rm{s}}^{\rm{2}}}}}} \right) \times 5\;{\rm{m}}\\ {v_1} &= 7.972\;{{\rm{m}} {\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. } {\rm{s}}} \end{align*} {/eq}

The expression for the time when the box is half way down the ramp is given as:

{eq}{t_1} = \dfrac{{{v_1} - {u_1}}}{a} {/eq}

Substitute the value in the above expression.

{eq}\begin{align*} {t_1} &= \dfrac{{7.972\;{{\rm{m}} {\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. } {\rm{s}}} - 6.324\;{{\rm{m}} {\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. } {\rm{s}}}}}{{2.356\;{{\rm{m}} {\left/ {\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right. } {{{\rm{s}}^{\rm{2}}}}}}}\\ {t_1} &= 0.7\;{\rm{s}} \end{align*} {/eq}

After the box come to the half way down the ramp, it collides and sticks to another box. The momentum is conserved before and after collision. The expression for the momentum is given as:

{eq}\begin{align*} {p_i} &= {p_f}\\ {m_1}{v_1} + {m_2}{u_2} &= \left( {{m_1} + {m_2}} \right){v_f} \end{align*} {/eq}

Substitute the value in the above equation.

{eq}\begin{align*} 2\;{\rm{kg}} \times 7.972\;{{\rm{m}} {\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. } {\rm{s}}} + 3\;{\rm{kg}} \times {\rm{0}} &= \left( {2\;{\rm{kg}} + 3\;{\rm{kg}}} \right){v_f}\\ {v_f} &= 3.188\;{{\rm{m}} {\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. } {\rm{s}}} \end{align*} {/eq}

Half the distance of the ramp, both the box will travel together. The expression for the distance is given as:

{eq}L = {v_f}t + \dfrac{1}{2}a{t^2} {/eq}

Substitute the value in the above equation.

{eq}\begin{align*} 5 &= 3.188 \times t + \dfrac{1}{2} \times 2.356\; \times {t^2}\\ 3.188\; \times {t^2} + 5.058\;t - 10 &= 0\\ t &= \dfrac{{ - 5.058 + \sqrt {{{5.058}^2} - 4 \times 3.188 \times \left( { - 10} \right)} }}{{2 \times 3.188}}\\ t &= 1.14\;{\rm{s}}\; \end{align*} {/eq}

The expression for the time taken to reach to the bottom is given as:

{eq}{t_b} = t + {t_1} {/eq}

Substitute the value in the above equation.

{eq}\begin{align*} {t_b} &= 1.14\;{\rm{s}}\; + 0.7\;{\rm{s}}\\ {t_b} &= 1.84\;{\rm{s}} \end{align*} {/eq}

Thus, the time taken to reach to the bottom of the ramp is {eq}1.84\;{\rm{s}} {/eq}.

#### Learn more about this topic:

from

Chapter 17 / Lesson 11In this lesson, you'll have the chance to practice using the spring constant formula. The lesson includes four problems of medium difficulty involving a variety of real-life applications.