Chapter 5.3, Problem 39PP

To determine

Acceleration of S.

Answer to Problem 39PP

  $a=5.2{\text{m/s}}^2$

Explanation of Solution

Given:

Mass of S, $m=45\text{kg}$

Angle of inclination, $\theta =45°$

Coefficient of friction, $\mu =0.25$

Formula Used:

From Newton’s second law of motion,

  $F=ma$

Where, $F$ is force, $m$ is mass and $a$ is acceleration of object.

Calculation:

The free body diagram of Stacie is,

  

As Stacie’s motion is downward. Therefore, force of friction $F_f$ will be upward.

Here, $F_N=F_g\cos \theta$ ( $F_N$ is normal force)

Now, net force in the direction of motion is, $F_{g\parallel }-F_f$

Here, $F_{g\parallel }$ is Stacie’s weight parallel to slide.

So, from Newton’s second law

  $\begin{array}{l}{F}_{g\parallel }-F_f=ma\\ a=\frac{F_{g\parallel }-F_f}{m}\\ \because F_{g\parallel }=mg\sin \theta \text{and }{F}_f={\mu }_k{F}_N\\ a=\frac{mg\sin \theta -{\mu }_k{F}_N}{m}\\ \because F_N=mg\cos \theta \\ a=\frac{mg\sin \theta -{\mu }_{k}mg\cos \theta }{m}\\ \Rightarrow a=g\sin \theta -{\mu }_{k}g\cos \theta \end{array}$

On putting the values,

  $\begin{array}{l}a=9.8\times \sin {45}^o-\left(0.25\times 9.8\times \cos {45}^o\right)\\ \Rightarrow a=5.2{\text{m/s}}^2\end{array}$

Conclusion:

Therefore, the acceleration of S is $a=5.2{\text{m/s}}^2$ .