91Ó°ÊÓ

In the absence of friction, there is no energy loss, and an object takes the same time going up as coming down.

Given data:

The mass of the block is $m=7.0\mathrm{kg}$.

The angle of incline is $\mathrm{θ}=22°$.

The initial speed of the block along the x-axis is $v_{\mathrm{i}}=-4.5\frac{\mathrm{m}}{\mathrm{s}}$.The final speed of the block is $v_{\mathrm{f}}=0$.

Assumption:

Assume that it goes $\mathrm{Δ}x$distance before coming to rest.

Let it take ttime to reach the highest point.

Part (a)

The normal force acting on the block is $N=mg\cos \mathrm{θ}$.

The net force on the block is

$F=mg\sin \mathrm{θ}$.

Then, the acceleration of the block along the x-axis is

$\begin{array}{c}a=\frac{mg\sin \mathrm{θ}}{m}\\ =g\sin \mathrm{θ}\end{array}$

Now,

$\begin{array}{c}{v}_{\mathrm{f}}^2=v_{\mathrm{i}}^2+2a\mathrm{Δ}x\\ v_{\mathrm{f}}^2=v_{\mathrm{i}}^2+2g\sin \mathrm{θ}\mathrm{Δ}x\\ 0^2={\left(-4.5\frac{\mathrm{m}}{\mathrm{s}}\right)}^2+2×9.80\frac{\mathrm{m}}{{\mathrm{s}}^2}×\sin {22}^o×\mathrm{Δ}x\\ \mathrm{Δ}x=-2.76\mathrm{m}\end{array}$

Hence, the block moves 2.76 m up the plane.

Part (b)

Now, for the upward motion,

$\begin{array}{c}{v}_{\mathrm{f}}=v_{\mathrm{i}}+at\\ v_{\mathrm{f}}=v_{\mathrm{i}}+g\sin \mathrm{θ}t\\ 0=-4.5\frac{\mathrm{m}}{\mathrm{s}}+\left(9.80\frac{\mathrm{m}}{{\mathrm{s}}^2}×\sin {22}^o\right)t\\ t=1.23\mathrm{s}\end{array}$

As there is no friction, the block takes the same time to return at the same position.

Then, the total time to come back to the starting position is

$\begin{array}{c}T=2t\\ =2×1.23\mathrm{s}\\ =2.46\mathrm{s}\end{array}$

Hence, it takes 2.46 s to come back to the starting position.