91影视

The given data can be listed below as:

• The mass of the box is $m=6.00\text{鈥塳驳}$.
• The angle of inclination of the box is $胃={37.0}^{鈭�}$.
• The kinetic frictional coefficient is${渭}_k=0.30$ .
• The acceleration of the box is $a=3.60{\text{鈥尘/蝉}}^{\text{2}}$.

The force is described as an agent which is responsible for changing the state of motion of an object. The force exerted by an object is equal to mass and the acceleration of that object.

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

According to the free body diagram, it can be identified that the summation of the force acting in the$x$and in the$y$direction is zero.

The equation of the force acting in the $x$direction is expressed as:

$F\cos 胃鈭�{渭}_{k}N鈭�W\sin 胃=ma$ 鈥�(1)

Here,$F$is the horizontal force,$胃$is the angle of inclination of the box,${渭}_k$is the kinetic frictional coefficient, $N$is the normal force,$W$is the weight of the box, $m$is the mass and is the acceleration of the box.

The equation of the force acting in the$x$direction is expressed as:

$F\sin 胃+W\cos 胃=N$

Substitute the value of the above equation in the equation (i).

$\begin{array}{c}F\cos 胃鈭�{渭}_k(F\sin 胃+W\cos 胃)鈭�W\sin 胃=ma\\ F\cos 胃鈭�{渭}_{k}F\sin 胃+{渭}_{k}W\cos 胃鈭�W\sin 胃=ma\\ F(\cos 胃鈭�{渭}_k\sin 胃)=ma+{渭}_{k}W\cos 胃+W\sin 胃\\ F=\frac{ma+{渭}_{k}W\cos 胃+W\sin 胃}{(\cos 胃鈭�{渭}_k\sin 胃)}\end{array}$

Substitute$mg$for$W$in the above equation.

$F=\frac{ma+{渭}_{k}mg\cos 胃+mg\sin 胃}{(\cos 胃鈭�{渭}_k\sin 胃)}$

Here,$g$ is the acceleration due to gravity.

Substitute the values in the above equation.

$\begin{array}{c}F=\frac{\left(6.00\text{鈥塳驳}\right)\left(3.60{\text{鈥尘/蝉}}^{\text{2}}\right)+\left(0.30\right)\left(6.00\text{鈥塳驳}\right)\left(9.8{\text{鈥尘/蝉}}^{\text{2}}\right)\cos {37.0}^{鈭�}+\left(6.00\text{鈥塳驳}\right)\left(9.8{\text{鈥尘/蝉}}^{\text{2}}\right)\sin {37.0}^{鈭�}}{(\cos {37.0}^{鈭�}鈭�\left(0.30\right)\sin {37.0}^{鈭�})}\\ =\frac{\left(21.6{\text{鈥塳驳.m/s}}^{\text{2}}\right)+\left(17.64{\text{鈥塳驳.m/s}}^{\text{2}}\right)\left(0.79\right)+\left(58.8{\text{鈥塳驳.m/s}}^{\text{2}}\right)\left(0.601\right)}{(\left(0.79\right)鈭�\left(0.30\right)\left(0.601\right))}\\ =\frac{\left(21.6{\text{鈥塳驳.m/s}}^{\text{2}}\right)+(13.93\text{鈥塳驳}鈰�{\text{m/s}}^{\text{2}})+\left(35.33{\text{鈥塳驳.m/s}}^{\text{2}}\right)}{(\left(0.79\right)鈭�\left(0.1803\right))}\\ =115\text{鈥塏}\end{array}$

Thus, the required horizontal force is$115\text{鈥塏}$ .