91Ó°ÊÓ

The given data can be listed below as:

• The mass of the box is m = 8.00 kg.
• The angle of inclination of the ramp with the horizontal is $\mathrm{θ}=33.0°$.
• The kinetic frictional coefficient is${\mathrm{μ}}_k=0.300$.
• The applied horizontal force is F = 26.0 N .
• The time taken by the box to reach the required speed is t = 2.00 s .

The force applied by an object is directly proportional to the acceleration and the mass of that object. Furthermore, the force applied in also directly proportional to the change in the velocity of that object.

The free body diagram of the box has been drawn below:

According to the free body diagram, there are four forces acting on the box such as the normal force n , the frictional force $f_x$, the horizontal force F and the weight of the box .

Splitting all the forces into components along X and Y-axis and then calculate the net force acting along the X-axis and Y-axis separately.

The relation for calculating net force along Y-axis is given as:

$n=mg\cos \mathrm{θ}-F\sin \mathrm{θ}$ …(¾±)

Here, n is the normal force, m is the mass of the box, g is the acceleration due to gravity, $\mathrm{θ}$is the angle subtended by the box and F is the horizontal force.

The relation for calculating net force along X-axis is given as:

$mg\sin \mathrm{θ}+F\cos \mathrm{θ}-f_k=ma$ …(¾±¾±)

Comparing equations (1) and (3).

$$\begin{gathered} \mathrm{mg}\mathrm{²õ¾\pm ²Ôθ}+\mathrm{F}\mathrm{³¦´Ç²õθ}-{\mathrm{μ}}_{\mathrm{k}}\left(\mathrm{mg³¦´Ç²õθ}-\mathrm{F²õ¾\pm ²Ôθ}\right)=\mathrm{ma} \\ \mathrm{mg}\mathrm{²õ¾\pm ²Ôθ}+\mathrm{F}\mathrm{³¦´Ç²õθ}-{\mathrm{μ}}_{\mathrm{k}}\mathrm{mg³¦´Ç²õθ}+{\mathrm{μ}}_{\mathrm{k}}\mathrm{F²õ¾\pm ²Ôθ}=\mathrm{ma} \\ \mathrm{a}=\frac{\mathrm{mg}\mathrm{²õ¾\pm ²Ôθ}+\mathrm{F}\mathrm{³¦´Ç²õθ}-{\mathrm{μ}}_{\mathrm{k}}\mathrm{mg³¦´Ç²õθ}+{\mathrm{μ}}_{\mathrm{k}}\mathrm{F²õ¾\pm ²Ôθ}}{\mathrm{m}} \end{gathered}$$

Substitute the given values in the above equation.

role="math" localid="1664884587725" $$\begin{gathered} \mathrm{a}=\frac{\left(\begin{array}{c}\left(8.00\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\sin 33.0°+\left(26.0\mathrm{N}\right)\cos 33.0°\\ -\left(0.300\right)\left(8.00\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\cos 33.0°+\left(0.300\right)\left(26.0\mathrm{N}\right)\sin 33.0°\end{array}\right)}{\left(8.00\mathrm{kg}\right)} \\ =\frac{\left(\begin{array}{c}\left(78.4\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2\right)\left(0.54\right)+\left(26.0\mathrm{N}\right)\left(0.83\right)-\left(0.300\right)\left(78.4\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2\right)\left(0.83\right)\\ +\left(0.300\right)\left(26.0\mathrm{N}\right)\left(0.54\right)\end{array}\right)}{\left(8.00\mathrm{kg}\right)} \\ =\frac{\left(42.336\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2\right)+\left(21.58\mathrm{N}\right)-\left(19.52\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2\right)+\left(4.2\mathrm{N}\right)}{\left(8.00\mathrm{kg}\right)} \\ =\frac{\left(22.81\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2\right)+\left(25.78\mathrm{N}\right)}{\left(8.00\mathrm{kg}\right)} \end{gathered}$$

Hence, further as:

$$\begin{gathered} \mathrm{a}=\frac{\left(22.81\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2\right)+\left(25.78\mathrm{N}\right)}{\left(8.00\mathrm{kg}\right)} \\ =\frac{\left(22.81\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2\right)+\left(25.78\mathrm{N}×{\displaystyle \frac{1\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2}{1\mathrm{N}}}\right)}{\left(8.00\mathrm{kg}\right)} \\ =\frac{\left(48.59\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2\right)}{\left(8.00\mathrm{kg}\right)} \\ =6.07\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

The equation of the final velocity of the box is expressed as:

v = u + at

Here, v is the final and u is the initial velocity of the box. t is the time taken by the box to reach that speed.

As initially the box was at rest, then the initial velocity of the box is zero.

Substitute the values in the above equation.

$$\begin{gathered} \mathrm{v}=0+\left(6.07\mathrm{m}/{\mathrm{s}}^2\right)\left(2\mathrm{s}\right) \\ =\left(6.07\mathrm{m}/{\mathrm{s}}^2\right)\left(2\mathrm{s}\right) \\ =12.14\mathrm{m}/\mathrm{s} \end{gathered}$$

Thus, the speed of the box is 12.14 m/s.