91Ó°ÊÓ

(a) Let's draw a free body diagram when the system is accelerating.

It is crucial to state that there is no friction, since a friction force, due to the discontinuity at the maximum , would not allow for a continuous measurement spectrum

Having drawn the free body diagram, we can project to the vertical and horizontal axes and have:

$n\cos \mathrm{θ}-F_G=0⇒n=\frac{F_G}{\cos \mathrm{θ}}=\frac{mg}{\cos \mathrm{θ}}$

$n\sin \mathrm{θ}=ma⇒a=\frac{n\sin \mathrm{θ}}{m}$

Substituting, we have

$a=\frac{\frac{mg}{\cos \mathrm{θ}}\sin \mathrm{θ}}{m}=g\tan \mathrm{θ}$

The key to solving is realizing that the tangent of our angle $\tan \mathrm{θ}$is equal to the slope of the line tangent to the $y=x^2$urve at the same point. The latter is easily determined, as one remembers from calculus, as

Combining, we have

Please note that we can't perform dimensional analysis on this result.

the acceleration will be

$\text{(b) If}x=0.2$

$a=2·9.8·0.2=3.92\mathrm{m}/{\mathrm{s}}^2$

If you are not sure why the angle between $n$and the vertical axis is equal to the angle between the line tangent to the curve and the $x\text{- axis,}$consider the close-up below.

${\mathrm{θ}}_1+{\mathrm{γ}}_1=90°$Since $n$is perpendicular to the surface and the line is tangent to this surface. Therefore we have ${\mathrm{γ}}_1=90°-{\mathrm{θ}}_1$. On the other hand, ${\mathrm{γ}}_1={\mathrm{γ}}_2$as opposite angles. But ${\mathrm{θ}}_2=90°-{\mathrm{γ}}_2$since they are both in the same right triangle. Therefore,

${\mathrm{θ}}_1={\mathrm{θ}}_2$

In the first sketch both angles were denoted by