91影视

Mass of the wood box: $2.0\mathrm{kg}$.

The angle of the force applied: ${45}^{鈭�}$.

Coefficient of friction between wood and wood : localid="1647680350969" $0.20$.

A figure:

$\begin{array}{r}F=mg\\ F_f=渭N=渭F_h\end{array}$

The force is applied at an angle of ${45}^{鈭�}$We will dissolve the applied force into horizontal and vertical components as

$\begin{array}{r}{F}_h=F\cos 鈦�{45}^{鈭�}\\ F_v=F\sin 鈦�{45}^{鈭�}\end{array}$

According to the free body diagram.

Now,

There are three forces acting at a time on the wood box. These forces are

1. Gravitation force $(F=mg)鈰�$downward

2. Vertical component of the force $(F_v=F\sin 鈦�{45}^{鈭�})鈰�$upward

3. Frictional force (in opposite of motion, $\left(F_f\right)$-upward

As, the box is sliding downward with constant speed. So

$\begin{array}{rl}& F_v+F_f鈭�mg=0\\ & mg=F_v+F_f\\ & mg=F\sin 鈦�{45}^{鈭�}+渭N\\ & mg=F\sin 鈦�{45}^{鈭�}+渭F\cos 鈦�{45}^{鈭�}\\ & mg=F(\sin 鈦�{45}^{鈭�}+渭\cos 鈦�{45}^{鈭�})\\ & 2\mathrm{kg}脳9.8\mathrm{m}/{\mathrm{s}}^2=F(0.707+0.2脳0.707)\\ & \frac{19.6}{0.8434}=F=23.10\mathrm{N}\end{array}$

$23.10N$Force should be applied to the box to slide down at constant speed.

$23.10N$Force should be applied to the box to slide down at constant speed.