91影视

Weight,${}^{\mathrm{W}=80\mathrm{N}}$

Coefficient of static friction,${}^{{\mathrm{渭}}_{\mathrm{s}=0.25}}$

Coefficient of kinetic friction,${}^{{\mathrm{渭}}_{\mathrm{k}}=0.15}$

Inclined angle,${}^{\mathrm{胃}={20}^0}$

The problem is based on Newton鈥檚 second law of motion which states that the rate of change of momentum of a body is equal in both magnitude and direction of the force acting on it. Use the Newton's 2nd law of motion along vertical and horizontal direction.

Formula:

$F_{net}=鈭�ma$

where,Fis the net force,mis mass andais an acceleration.

Free body Diagram of sled over the inclined plane:

First, from the weight, calculate the mass of the system,

$$\begin{gathered} W=mg \\ m=\frac{W}{g} \\ =\frac{80\mathrm{N}}{9.81} \\ =8.2\mathrm{kg} \end{gathered}$$

By using Newton鈥檚 2nd law along vertical direction (along y),

$鈭�F_y=m{a}_y$

since, block is not moving upward, $a_y=0$

$$\begin{gathered} N-F_g\cos \left(胃\right)=0 \\ N=F_g\cos \left(胃\right) \end{gathered}$$

Relation between static frictional force and normal force is,

$f_s={渭}_{s}N=N_s{F}_g\cos \left(胃\right)$ (i)

When sled is moving down frictional force is upward,

Now, applying Newton鈥檚 2nd along the x direction, for fig. a,

$鈭�F_s=m{a}_x$

since, block is not moving, localid="1654149316961" $a_x=0$

$$\begin{gathered} F+f_s-F_g\sin \left(胃\right)=0 \\ F=f_g\sin \left(胃\right)-f_s \end{gathered}$$

By using equation (i) in above equation,

$$\begin{gathered} F=F_g\sin \left(胃\right)-{渭}_s{F}_g\cos \left(胃\right) \\ F=F_g(\sin \left(胃\right)-{渭}_s\cos \left(胃\right)) \\ =mg(\sin \left(20\right)-(0.25\left)\cos \left(20\right)\right) \\ =8.6N \end{gathered}$$

Thus, the least magnitude of force ${}^F$that prevents the sled moving down is ${}^{8.6\mathrm{N}}$

To find the F to start the sled moving up is, again we have to use Newton鈥檚 2nd law along y and x direction, but in this case as the sled moving up frictional force is downward.So, refer to fig. b.

Along vertical direction (y),

$鈭�F_y=m{a}_y$

since block is not moving upward,$a_y=0$

$$\begin{gathered} N-F_g\cos \left(胃\right)=0 \\ N=F_g\cos \left(胃\right) \end{gathered}$$

Then static frictional force,

$f_s={渭}_{s}N={渭}_s{F}_g\cos \left(胃\right)$

Along horizontal direction (x),

$鈭�F_x=m{a}_x$

since, block is not moving, $a_x=0$

$$\begin{gathered} F-f_s-F_g\sin \left(胃\right)=0 \\ F=F_g\sin \left(胃\right)+f_s \\ =F_g(\sin 胃+{渭}_s\cos 胃) \\ =46N \end{gathered}$$

Hence, the minimum magnitude of F that will start the sled moving up the plane is ${}^{46\mathrm{N}}$.

To find ${}^F$to move the sled up the plane at constant velocity, use kinetic frictional force,

$$\begin{gathered} f_k={渭}_{k}N \\ ={渭}_k{F}_g\cos \left(胃\right) \end{gathered}$$

Along horizontal direction (x),

$鈭�F_x=m{a}_x$

since, block is moving with constant velocity, $a_x=0$

$$\begin{gathered} F-f_k-F_g\sin \left(胃\right)=0 \\ F=F_g\sin \left(胃\right)+f_k \\ =F_g(\sin 胃+{渭}_k\cos 胃) \\ =39N \end{gathered}$$

Hence, the value of F, that is required to move the sled up the plane at constant velocity is${}^{39\mathrm{N}}$