A block is given an initial speed of 4.5 m/s up a 22.0° plane, as shown in Fig. 4–59. (a) How far up the plane will it go? (b) How much time elapses before it returns to its starting point? Ignore the friction.

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 $\theta =22°$.

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 $\Delta x$distance before coming to rest.

Let it take ttime to reach the highest point.

Part (a)

The normal force acting on the block is $N=mg\cos \theta$.

The net force on the block is

$F=mg\sin \theta$.

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

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

Now,

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

Hence, the block moves 2.76 m up the plane.

Part (b)

Now, for the upward motion,

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

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

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

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