Question

A 300 g block is dropped onto a relaxed vertical spring that has a spring constant of k = 2.1 N/cm. The block becomes attached to the spring and compresses the spring 12 cm before momentarily stopping.

(a) While the spring is being compressed, what work is done on the block by the gravitational force on it?
J

(b) What work is done on the block by the spring force while the spring is being compressed?
J

(c) What is the speed of the block just before it hits the spring? (Assume that friction is negligible.)
m/s

(d) If the speed at impact is doubled, what is the maximum compression of the spring?
m

Answer 1

Concepts and reason

The concept used in this problem are potential energy of a spring and the law of conservation of energy.

First convert the units and use the work done by a force to calculate work done by gravitational force. Then, calculate the work done by the spring on block while it is being compressed by calculating the potential energy of spring under compression. Then, calculate Kinetic energy of the block before the impact by using law of conservation of energy. Then, find the initial value of kinetic energy if the speed of block at impact is doubled and express new kinetic energy in terms of old kinetic energy. Finally, use energy conservation equation to find new value of compression $d'$ .

Fundamentals

The force of gravitation is the force on an object by the earth. It is given as,

${F_{\rm{g}}} = mg$

Here, $m$ is mass and $g$ is the acceleration due to gravity.

Work- Work is defined as the product of force and the displacement along the direction of force. This can be expressed as follows:

$W = Fd\cos \theta$

Here, $W$ is work, $F$ is force, $d$ is displacement and $\theta$ is the angel between force and displacement.

Potential energy of a spring- This is the defined as the amount of work required to compress or stretch the spring. If the spring has the spring constant equal to $k$ and the change in its length is $x$ then potential energy stored in the spring is expressed as follows:

${U_{spring}} = \frac{1}{2}k{x^2}$

Law of conservation of energy-The law states that total energy of the system remains conserved unless acted upon by an external force.

Convert the unit of mass in SI unit system as follows:

$\begin{array}{c}\\m = 300{\rm{ g}}\\\\ = \left( {{\rm{300 g}}} \right)\left( {\frac{{1{\rm{ kg}}}}{{1000{\rm{ g}}}}} \right){\rm{ }}\\\\ = 0.3{\rm{ kg}}\\\end{array}$

Convert the unit of displacement in SI unit system as follows:

$\begin{array}{c}\\d = 12{\rm{ cm}}\\\\ = \left( {12{\rm{ cm}}} \right)\left( {\frac{{1{\rm{ m}}}}{{100{\rm{ cm}}}}} \right)\\\\ = 0.12{\rm{ m}}\\\end{array}$

Calculate the gravitational force on the block as follows:

Substitute ${\rm{0}}{\rm{.3 kg}}$ for $m$ and ${\rm{9}}{\rm{.8 m}} \cdot {{\rm{s}}^{ - 2}}$ for $g$ in the equation ${F_g} = mg$ to calculate the gravitational force.

$\begin{array}{c}\\{F_g} = \left( {{\rm{0}}{\rm{.3 kg}}} \right)\left( {{\rm{9}}{\rm{.8 m}} \cdot {{\rm{s}}^{ - 2}}} \right){\rm{ }}\\\\ = 2.94{\rm{ kg}} \cdot {\rm{m}} \cdot {{\rm{s}}^{ - 2}}\\\\ = 2.94{\rm{ N}}\\\end{array}$

Calculate the work done by gravitational force as follows:

Substitute ${\rm{2}}{\rm{.94 N}}$ for ${F_g}$ , ${\rm{0}}{\rm{.12 m}}$ for $d$ and $0^\circ$ for $\theta$ in the equation ${W_g} = {F_g}d\cos \theta$ to calculate work done by gravitational force.

$\begin{array}{c}\\{W_g} = \left( {{\rm{2}}{\rm{.94 N}}} \right)\left( {{\rm{0}}{\rm{.12 m}}} \right){\rm{cos}}\left( {{\rm{0}}^\circ } \right){\rm{ }}\\\\ = 0.3528{\rm{ J}}\\\\ = 0.35{\rm{ J}}\\\end{array}$

Hence, the work done on block by gravitational force is $0.35{\rm{ J}}$ .

Calculate potential energy of the spring when it is compressed by amount $d$ as follows:

Convert the unit of spring constant in SI form.

$\begin{array}{c}\\k = 2.1{\rm{ N}}/{\rm{cm}}\\\\ = \left( {2.1\frac{{\rm{N}}}{{{\rm{cm}}\left( {\frac{{1{\rm{ m}}}}{{100{\rm{ cm}}}}} \right)}}} \right)\\\\ = 210{\rm{ N}} \cdot {{\rm{m}}^{ - 1}}\\\end{array}$

The force exerted by the spring on the block is opposite to the displacement of the block.

The work done by the spring will be equal to negative of the work done on the spring by the block.

This can be expresses as follows:

${W_{spring}} = - {U_{spring}}$

Substitute $\frac{1}{2}k{d^2}$ for ${U_{spring}}$ in the above equation ${W_{spring}} = - {U_{spring}}$ .

${W_{spring}} = - \frac{1}{2}k{d^2}$

Substitute $210{\rm{ N}} \cdot {{\rm{m}}^{ - 1}}$ for $k$ , and ${\rm{0}}{\rm{.12 m}}$ for $d$ in the equation ${W_{spring}} = - \frac{1}{2}k{d^2}$ to calculate the work done by the spring.

$\begin{array}{c}\\{W_{spring}} = - \left( {\frac{1}{2}} \right)\left( {210{\rm{ N}} \cdot {{\rm{m}}^{ - 1}}} \right){\left( {{\rm{0}}{\rm{.12 m}}} \right)^2}\\\\ = - 1.512{\rm{ J}}\\\\ = - {\rm{1}}{\rm{.51 J}}\\\end{array}$

Hence, work done on the block by spring is $- {\rm{1}}{\rm{.51 J}}$ .

Calculate Kinetic energy of the block as follows:

Kinetic energy of the block when it momentarily comes to rest is 0 as there is no motion.

The total stored energy of the spring is the result of sum of change in kinetic energy and potential energy of the block. This can be written as follows;

${U_{spring}} = \left( {K.{E_i} - K.{E_f}} \right) + mg \cdot d$

Here, $K.{E_i}$ is the initial kinetic energy of the block before the impact $K.{E_f}$ is the final kinetic energy of the block when it momentarily comes to rest.

Substitute 1.512 J for ${U_{spring}}$ , 0 for $K.{E_f}$ , ${\rm{2}}{\rm{.94 N}}$ for ${F_g}$ , ${\rm{0}}{\rm{.12 m}}$ for $d$ in the equation ${U_{spring}} = \left( {K.{E_i} - K.{E_f}} \right) + mgd$ to calculate the initial kinetic energy.

$\begin{array}{c}\\1.512{\rm{ J}} = K.{E_i} + 0.3528{\rm{ J}}\\\\K.{E_i} = \left( {1.512 - 0.3528} \right){\rm{ J}}\\\\ = 1.1592{\rm{ J}}\\\end{array}$

Calculate speed of the block just before it hits the spring as follows:

Substitute ${\rm{0}}{\rm{.3 kg}}$ for $m$ and $1.1592{\rm{ J}}$ for $K.{E_i}$ in the equation ${v_i} = \sqrt {\frac{{2K.{E_i}}}{m}}$ to calculate the initial velocity.

$\begin{array}{c}\\{v_i} = \sqrt {\frac{{2\left( {1.1592{\rm{ J}}} \right)}}{{{\rm{0}}{\rm{.3 kg}}}}} \\\\ = 2.78{\rm{ m/s}}\\\end{array}$

Hence, the speed of the block just before it hits the spring is $2.78{\rm{ m/s}}$ .

Take initial speed of block at impact to be $v$ and new speed to be $v'$ .

The expression of new kinetic energy is as follows;

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

Substitute $2v$ for $v'$ in the equation $K.E' = \frac{1}{2}m{v'^2}$ .

$\begin{array}{c}\\K.E' = \frac{1}{2}m{\left( {2v} \right)^2}\\\\ = 4\left( {\frac{1}{2}m{v^2}} \right)\\\end{array}$

Substitute $1.1592{\rm{ J}}$ for $\left( {\frac{1}{2}m{v^2}} \right)$ in the equation $K.E' = 4\left( {\frac{1}{2}m{v^2}} \right)$ .

$\begin{array}{c}\\K.E' = 4\left( {1.1592{\rm{ J}}} \right)\\\\ = 4.6368{\rm{ J}}\\\end{array}$

By law of conservation of energy, sum of change in kinetic and potential energy is equal to change in potential energy of the spring during compression.

${U'_{spring}} = K.E' + mgd'$

Here, $d'$ is the new compressed length.

Calculate the potential energy of spring under new compression $d'$ as follows:

${U'_{spring}} = \frac{1}{2}k{\left( {d'} \right)^2}$

Substitute $210{\rm{ N}} \cdot {{\rm{m}}^{ - 1}}$ for $k$ in the equation ${U'_{spring}} = \frac{1}{2}k{\left( {d'} \right)^2}$ and solve for potential energy of the spring.

$\begin{array}{c}\\{{U'}_{spring}} = \frac{1}{2}\left( {210{\rm{ N}} \cdot {{\rm{m}}^{ - 1}}} \right){\left( {d'} \right)^2}\\\\ = 105{\left( {d'} \right)^2}{\rm{ N}} \cdot {{\rm{m}}^{ - 1}}\\\end{array}$

Substitute $105{\left( {d'} \right)^2}$ for ${U'_{spring}}$ , $4.6368{\rm{ J}}$ for $K.E'$ , $0.3{\rm{ kg}}$ for $m$ , and $9.8{\rm{ m}} \cdot {{\rm{s}}^{ - 2}}{\rm{ for }}g$ in the equation ${U'_{spring}} = K.E' + mgd'$ and solve for $d'$ .

$\begin{array}{l}\\105{\left( {d'} \right)^2}\;{\rm{N}} \cdot {{\rm{m}}^{ - 1}} = 4.6368{\rm{ J}} + \left( {0.3{\rm{ kg}}} \right)\left( {9.8{\rm{ m}} \cdot {{\rm{s}}^{ - 2}}} \right)d'\\\\105{\left( {d'} \right)^2}{\rm{ N}} \cdot {{\rm{m}}^{ - 1}} = 4.6368{\rm{ N}} \cdot {\rm{m }} + 2.94d'{\rm{ kg}} \cdot {\rm{m}} \cdot {{\rm{s}}^{ - 2}}\\\\105{\left( {d'} \right)^2}{\rm{ N}} \cdot {{\rm{m}}^{ - 1}} = 4.6368{\rm{ N}} \cdot {\rm{m}} + 2.94d'{\rm{ N}}\\\end{array}$

Rearrange the equation as follows:

$105{\left( {d'} \right)^2} - 2.94d' - 4.6368 = 0$

Solve the quadratic to obtain the value of $d'$ as follows:

Substitute $d'$ for $x$ , $105{\rm{ N}} \cdot {{\rm{m}}^{ - 1}}$ for $a$ , $- 2.94{\rm{ N}}$ for $b$ and $- 4.6368{\rm{ J}}$ for $c$ in the equation $x = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ to calculate the $d'$ .

$\begin{array}{c}\\d' = \frac{{ - \left( { - 2.94{\rm{ N}}} \right) \pm \sqrt {{{\left( { - 2.94{\rm{ N}}} \right)}^2} - 4\left( {105{\rm{ N}} \cdot {{\rm{m}}^{ - 1}}} \right)\left( { - 4.6368{\rm{ N}} \cdot {\rm{m}}} \right)} }}{{2\left( {105{\rm{ N}} \cdot {{\rm{m}}^{ - 1}}} \right)}}\\\\ = 0.22{\rm{ m}}, - {\rm{0}}{\rm{.19 m}}\\\end{array}$

$d' = - 0.19{\rm{ m}}$ Indicates stretching of spring and is discarded.

Hence, the maximum compression of the spring is $d' = 0.22{\rm{ m}}$ .

Ans: Part a

The work done on block by gravitational force is $0.35{\rm{ J}}$ .

Part b

Work done on the block by spring is $- {\rm{1}}{\rm{.51 J}}$ .

Part c

The speed of the block just before it hits the spring is $2.78{\rm{ m/s}}$ .

Part d

The maximum compression of the spring is $0.22{\rm{ m}}$ .