Chapter 11, Problem 87A

To determine

Energy added by the spring.

Answer to Problem 87A

  $13.23\text{J}$ of energy was added by the spring.

Explanation of Solution

Given:

Mass of carts, $m_1=2\text{kg}$ and $m_2=1\text{kg}$

Initial speed of lab cart, $v=2.1\text{m/s}$

The 2 kg cart come to rest due to uncompressed spring.

Formula Used:

Kinetic energy is,

  $E_i=\frac{1}{2}m{v}^2$ …. (1)

Here, $m$ is mass of system and $v$ is velocity.

Momentum is,

  $p_i=mv$ …. (2)

Here, , $m$ is mass and $v$ is velocity

Calculation:

From (1) initial energy of the system is,

  $\begin{array}{c}{E}_i=\frac{1}{2}\times \left(2+1\right)\times {2.1}^2\\ =6.615\text{J}\end{array}$

From (2) initial momentum of the system is,

  $\begin{array}{c}{p}_i=\left(2+1\right)\times 2.1\\ =6.3\text{J}\end{array}$

As the 2 kg cart come to rest due to uncompressed spring. So from conservation of momentum, $p_i=p_f$

So,

  $\begin{array}{c}1\times v_f=6.3\\ v_f=6.3\text{m/s}\end{array}$

So, final energy of the system is

  $\begin{array}{c}{E}_f=\frac{1}{2}m{v}_f{}^2\\ =\frac{1}{2}\times 1\times {6.3}^2\\ =19.845\text{J}\end{array}$

Thus, change in energy or energy added by spring is

  $\begin{array}{c}\Delta E=19.845-6.615\\ =13.23\text{J}\end{array}$

Conclusion:

Therefore, $13.23\text{J}$ of energy was added by the spring.