91Ó°ÊÓ

We have given that:

The mass of package (\(m\)) = 0.200 kg.

Radius of arc of circular track (\(R\)) = 1.60 m.

Length (distance) of the horizontal path (\(s\))= 3.00 m.

Velocity of the package (particle) (\(v\)) = 4.80 m/s.

Let us take \(h = 0\) at point B and C level.

Consider the following cases:

Case 1) From point A to point B.

Taking A at one point and B at another point, we have,

\(\begin{aligned}{}{h_1} = R,\,\,{v_1} = 0\\{h_2} = 0,\,\,{v_2} = 4.80\end{aligned}\)

Case 2) From point B to point C.

Taking B at one point and C at another point, we have,

\({v_1} = 4.80,\,\,\,\,{v_2} = 0\)

Since there are other forces acting, the work-energy theorem is given by,

\({K_i} + {U_i} + W = {K_f} + {U_f}\,\,\, \cdots \cdots \left( 1 \right)\) .

Where the kinetic energy is given by,

\(K = \frac{1}{2}m{v^2}\,\, \cdots \cdots \left( 2 \right)\)

And the gravitational potential energy is given by,

\(U = mgh\,\,\, \cdots \cdots \left( 3 \right)\).

We know that the friction force and the resistance always act in the opposite direction to the direction of motion.

Therefore, the work done is given by,

\({W_f} = - {f_t} \cdot s\,\,\,\,\, \cdots \cdots \left( 4 \right)\)

The kinetic Friction force is,

\({f_k} = {\mu _k} \cdot n\,\,\,\, \cdots \cdots \left( 5 \right)\) , where \(n\) is the normal force

First, we calculate the energy quantities for the second case.

Now, substituting the values in \(\left( 2 \right)\), we get,

\(\begin{aligned}{}{K_1} = \frac{1}{2} \times \left( {0.20} \right) \times {\left( {4.80} \right)^2}\,\\{K_1} = 2.304J\end{aligned}\)

Since package eventually comes to rest, we have,

\({K_2} = 0\)

Also, the package moves horizontally on the 0 potential levels, we get,

\({U_1} = 0,\,\,{U_2} = 0\).

Applying Newton’s Second Law, to the package along the vertical direction, we get,

\(\sum {{F_y} = n - mg} = 0\)

\(\begin{aligned}{} \Rightarrow n = mg\\ \Rightarrow n = \left( {0.20} \right) \times \left( {9.8} \right) = 1.96N\end{aligned}\)

From \(\left( 5 \right)\), substituting this value, we get,

\({f_k} = 1.96{\mu _k}\,\,\)

Next, substituting these values in \(\left( 4 \right)\) with, we get,

\(W = {W_f} = - \left( {1.96{\mu _k}} \right) \cdot \left( {3.00} \right) = - 5.88{\mu _k}\)

Now, substituting all these work and energy values in \(\left( 1 \right)\), we get,

\(\begin{aligned}{}2.304 - 5.88{\mu _k} = 0\\ \Rightarrow 5.88{\mu _k} = 2.304\\ \Rightarrow {\mu _k} = \frac{{2.304}}{{5.88}}\\ \Rightarrow {\mu _k} = 0.40\end{aligned}\)

Hence, the coefficient of the kinetic friction on the horizontal surface is \({\mu _k} = 0.40\).

Now, we calculate the energy quantities for first case.

Substituting the values of \({h_1},\,\,m\) in \(\left( 3 \right)\), we get,\({U_1} = \left( {0.20} \right) \times \left( {9.8} \right) \times \left( {1.60} \right) = 3.136J\)

Since package starts at the rest, we have,

\({K_1} = 0\).

Also, the package ends at the 0 potential level, we get,

\({U_2} = 0\).

We have the final kinetic energy (\({K_1}\) in part (a)), then

\({K_2} = 2.304J\)

Substituting all these energy values in \(\left( 1 \right)\), we get,

\(\begin{aligned}{}0 + 3.136 + W = 2.304 + 0\\ \Rightarrow W = 2.304 - 3.136\\ \Rightarrow W = - 0.832J\end{aligned}\)

Since friction force is the only other than gravity and the normal force is always perpendicular to motion’s direction, therefore,

\({W_f} = W = - 0.832J\)

\(\therefore {W_f} = - 0.832J\)

Hence, the work done on the package by the friction is \({W_f} = - 0.832J\).