Chapter 5.2, Problem 29SSC

To determine

Initial speed of the card.

Answer to Problem 29SSC

  $1.28\text{ m/s}$

Explanation of Solution

Given:

Mass of card, $m=2.3\text{ g = 0}\text{.0023 kg}$

Coefficient of kinetic friction, ${\mu }_k=0.24$

Distance slides by card, $s=0.35\text{m}$

As the card finally stops therefore final velocity, v = 0

Calculation:

Acceleration due to gravity is given by,

  $g=9.8{\text{ m/s}}^2$

Now, substitute the value of ${\mu }_k$ , m and gand solve:

  $\begin{array}{l}F=0.24\times 0.0023\times 9.8\\ =5.4\times {10}^{-3}\text{ N}\end{array}$

Therefore, frictional force is equal to $5.4\times {10}^{-3}\text{ N}$

Now, work done by frictional force is equal to,

  $\begin{array}{l}W=-F\cdot s\\ =-5.4\times {10}^{-3}\times 0.35\\ =-1.89\times {10}^{-3}\text{ J}\end{array}$

Here, negative sign represent work is done in opposite direction

From work energy theorem work done is equal to change in kinetic energy.

Therefore,

  $\begin{array}{l}\frac{1}{2}m\left(v^2-u^2\right)=-1.89\times {10}^{-3}\\ \frac{1}{2}\times 0.0023\times \left(0^2-u^2\right)=-1.89\times {10}^{-3}\\ -u^2=-1643.47\times {10}^{-3}\\ u^2=1.643\\ u=1.28\text{ m/s}\end{array}$

Conclusion:

Therefore, initial speed of the card is $1.28\text{ m/s}$ .