Question

When an object slides on a surface, it encounters a resistance force called friction. This force has a magnitude of $\mathbf{\mu N}$ , where$\mathbf{\mu }$ the coefficient of kinetic friction and N isthe magnitude of the normal force that the surface applies to the object. Suppose an object of mass30 kgis released from the top of an inclined plane that is inclined $\mathbf{30}\mathbf{°}$to the horizontal (see Figure 3.11). Assume the gravitational force is constant, air resistance is negligible, and the coefficient of kinetic friction$\mathbf{\mu }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{2}$ . Determine the equation of motion for the object as it slides down the plane. If the top surface of the plane is 5 m long, what is the velocity of the object when it reaches the bottom?

Short answer

The velocity of the object when it reaches the bottom is$V\left(t\right)=5.66m/\sec$ and $x\left(t\right)=5.66t+c$.

Step-by-step solution

Find the value of velocity

There are two forces are written as,

$$\begin{gathered} F_1=mg \sin {30}^o \\ {F}_2=-\mu mg \cos {30}^o \end{gathered}$$

Now put the given values then;

$$\begin{gathered} m\frac{dv}{dt}=mgsin{30}^o-\mu mgcos{30}^o \\ \frac{dv}{dt}=gsin{30}^o-\mu gcos{30}^o \left(\mathrm{Puttingthevaluesof} \mathrm{g}=0.2\right) \\ \frac{dv}{dt}=3.207 \end{gathered}$$

Since$v\left(t\right)=V\left(x\left(t\right)\right)$,

So, the values are written as;

$$\begin{gathered} \frac{dv}{dt}=V\frac{dV}{dx} \\ V\frac{dV}{dx}=3.207 \end{gathered}$$

Find the value of velocity by limits.

Now, find the value of velocity then,

$$\begin{gathered} \int VdV=\underset{0}{\overset{5}{\int }}3.207dx \\ \frac{V^2}{2}-{\left.3.207x\right|}_0^5=0 \left(\mathbf{Integratewithlimits}\mathbf{0}\mathbf{to}\mathbf{5}\right) \\ {V}^2=32.07 \\ V\left(t\right)=5.66m/\sec \end{gathered}$$

Evaluate the equation of motion.

Now, for the value of x(t),

$$\begin{gathered} v=5.66 \\ \frac{dx}{dt}=5.66 \\ x \left(t\right)=5.66 t+c \end{gathered}$$

Therefore, the results are $x\left(t\right)=5.66 t+c$ and $v\left(t\right)=5.66 m/\sec$.