91Ó°ÊÓ

1. The initial speed of the block is $v_0=10\mathrm{m}/{\mathrm{s}}^2$
2. The angle of incline is$\mathrm{θ}=32.0°$

According to Newton’s second law, the net force acting on the object is equal to the product of mass and acceleration. The net force is also equal to the vector sum of all forces acting on the object. The acceleration of the object is the rate of change of velocity with respect to time. The acceleration can also be found using the kinematic equations if initial velocity, final velocity, and displacement are known.

Draw the free-body diagram of the block which is moving along the frictionless inclined plane. By using Newton’s second law motion find out the acceleration of the block. The block covers some distance and stops hence final velocity will be zero. By using kinematic equations, we can find the distance, time, and final speed of the block when it gets back to the bottom.

Formulae:

${\mathbf{F}}_{\mathbf{net}}\mathbf{}\mathbf{=}\mathbf{ma}$ (1)

${\mathit{v}}_{\mathbf{f}}^{\mathbf{2}}\mathbf{}\mathbf{=}{\mathit{v}}_{\mathbf{0}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{a}\mathbf{}\mathbf{d}$ (2)

${\mathbf{v}}_{\mathbf{f}}\mathbf{=}{\mathbf{v}}_{\mathbf{0}}\mathbf{+}\mathbf{at}$ (3)

$\mathbf{d}\mathbf{=}{\mathbf{v}}_{\mathbf{0}}\mathbf{t}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{at}}^{\mathbf{2}}$ (4)

According to the free body diagram, the block is moving up along a frictionless surface. The normal force N is acting in an upward direction and the gravitational force is acting in the downward direction with an angle $\mathrm{θ}$with the vertical axis. Hence $\mathrm{mg}.\cos \mathrm{θ}$and $\mathrm{mg}.\sin \mathrm{θ}$is are the vertical and horizontal components of mg as shown in the figure.

According to Newton’s second law,

$F_{net}=ma$

We can apply this law along the x-axis as

$$\begin{gathered} -\mathrm{mg}.\sin \mathrm{θ}=\mathrm{ma} \\ \mathrm{a}=-\mathrm{g}.\mathrm{²õ¾\pm ²Ôθ} \end{gathered}$$

We can use the sign convention according to the motion of an object. At the position at which the block stops, the final velocity v of the block is zero.

By using the third kinematical equation, we can find the distance covered by the block up the plane as,

$$\begin{gathered} v_f^2=v_0^2+2\mathrm{ad} \\ 0={\mathrm{v}}_0^2-2×\left(\mathrm{g}\sin \mathrm{θ}\right)+\mathrm{d} \\ {\mathrm{v}}_0^2=2×\left(\mathrm{g}\sin \mathrm{θ}\right)+\mathrm{d} \\ \mathrm{d}=\frac{{\mathrm{v}}_0^2}{2\mathrm{g}\sin \mathrm{θ}} \end{gathered}$$

Now, substitute the given values in the above equation.

$$\begin{gathered} \mathrm{d}=\frac{{\left(3.50\mathrm{m}/\mathrm{s}\right)}^2}{2×{\left(9.8\mathrm{m}/\mathrm{s}\right)}^2\sin \left(32°\right)} \\ =1.18\mathrm{m} \end{gathered}$$

Hence, the distance covered by the block up the plane is$1.18\mathrm{m}$.

By using the first kinematical equation, we can find the time taken by the block to stop.

$$\begin{gathered} v_f=v_0+\mathrm{at} \\ 0={\mathrm{v}}_0-\mathrm{g}.\mathrm{²õ¾\pm ²Ôθ}.\mathrm{t} \\ \mathrm{t}=\frac{{\mathrm{v}}_0}{\mathrm{g²õ¾\pm ²Ôθ}} \end{gathered}$$

Substitute the given values.

$$\begin{gathered} \mathrm{t}=\frac{3.50\mathrm{m}/\mathrm{s}}{\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)×\sin \left(32°\right)} \\ =0.674\mathrm{s} \end{gathered}$$

Hence, the time taken by the block to reach there is 0.674s

When the block gets back to the bottom, its displacement is zero. According to the third kinematical equation,

$$\begin{gathered} v^2=v_0^2+2\mathrm{ad} \\ ={\mathrm{v}}_0^2-2\mathrm{²\mu ³¦´Ç²õθ}×0 \end{gathered}$$

Substituting the values, we get.

$$\begin{gathered} v^2=v_0^2-2\mathrm{²\mu ³¦´Ç²õθ}×0 \\ {\mathrm{v}}^2={\left(3.50\mathrm{m}/\mathrm{s}\right)}^2 \\ \mathrm{v}=Â\pm 3.50\mathrm{m}/\mathrm{s} \end{gathered}$$$$\begin{gathered} v^2=v_0^2+2\mathrm{²\mu ³¦´Ç²õθ}×0 \\ {\mathrm{v}}^2={\left(3.50\mathrm{m}/\mathrm{s}\right)}^2 \\ \mathrm{v}=Â\pm 3.50\mathrm{m}/\mathrm{s} \end{gathered}$$

The block is going downward. Hence, the speed of the block is negative.