91Ó°ÊÓ

The mass of the skier (\(m\)) = 60.0 kg.

The slope of ski = 65.0 m.

Consider three stages:

force is \({W_f} = - 10.5\)kJ.

Taking \(h = 0\) at the bottom of the slope, we have,

\(\begin{aligned}{}{h_{\,i}} = 65,\,\,\,\,\,{v_i} = 0\\{h_f} = 0,\,\,\,\,\,\,\,{v_f} = \,{v_{\,2}}\end{aligned}\)

Then we have,

\({v_i} = {v_2},\,\,\,\,{v_f} = {v_3}\)

Then we have,

\({v_i} = {v_3},\,\,\,\,{v_f} = 0\)

Since there is other force than gravity, 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,

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

Kinetic Friction force is,

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

Let us calculate energy quantities.

Now, substituting the values of \({h_i},\,m\) in \(\left( 3 \right)\), we get,

\({U_i} = \left( {60} \right) \times \left( {9.8} \right) \times \left( {65} \right) = 38220\)J.

Since the skier starts from the rest, we have,

\({K_i} = 0\).

Also, she ends at \(h = 0\), therefore, we get,

\({U_f} = 0\).

Thus, substituting all these values of energy quantities in \(\left( 1 \right)\), we get,

\(\begin{aligned}{}38220 + 0 - 10.5 \times {10^3} = {K_f} + 0\\ \Rightarrow {K_f} = 27720J\end{aligned}\)

Using this value in \(\left( 2 \right)\), we will solve for \({v_2}\).

\(\begin{aligned}{}27720 = \frac{1}{2} \times \left( {60} \right) \times v_2^2\\ \Rightarrow {v_2} = \sqrt {\frac{{27720 \times 2}}{{60}}} \\ \Rightarrow {v_2} = 30.4\end{aligned}\)

Hence, the velocity while she is going at the bottom of the slope is \({v_2} = 30.4\)m/s.

Since she moves horizontally at \(h = 0\), so the work done gravity is zero.

Therefore, the equation \(\left( 1 \right)\) becomes,

\({K_i} + W = {K_f}\, \cdots \cdots \left( 6 \right)\)

By applying Newton’s Second Law, to the skier vertically, we get,

\(\begin{aligned}{}\sum {{F_y} = n - mg = 0} \\ \Rightarrow m = mg = 60 \times 9.8 = 588\end{aligned}\)

From \(\left( 5 \right)\), we get,

\({f_k} = 0.20 \times 588 = 117.6\)N.

Thus, total force acting on the skier along the patch is,

\({f_t} = {f_k} + f = 117.6 + 160 = 277.6\)N.

Substituting this value in \(\left( 4 \right)\), we get,

\(W = - \left( {277.6} \right) \times \left( {82} \right) = - 22763.2\)J.

We have initial kinetic energy (\({K_f}\)in part (a)), then

\({K_i} = 27720\)J.

Therefore, from \(\left( 6 \right)\), we get,

\(\begin{aligned}{}27720 - 22763.2 = {K_f}\\ \Rightarrow {K_f} = 4956.8J\end{aligned}\)

Substituting this value in \(\left( 2 \right)\) and solving for \({v_3}\), we get,

\(\begin{aligned}{}4956.8 = \frac{1}{2} \times \left( {60} \right) \times v_3^2\\ \Rightarrow {v_3} = \sqrt {\frac{{2 \times 4956.8}}{{60}}} \\ \Rightarrow {v_3} = 12.8\end{aligned}\)

Hence, the velocity after crossing the patch is \({v_3} = 12.8\)m/s.

Here she is still moving horizontally, we use equation \(\left( 6 \right)\) .

As she stops at a point, we get, \({K_f} = 0\).

We have initial kinetic energy (\({K_f}\) in part (b)), then,

\({K_i} = 4956.8\)J.

Putting all these values in \(\left( 6 \right)\), we get,

\(4956.8 + W = 0 \Rightarrow W = - 4956.8\)J.

Substituting the values in \(\left( 4 \right)\), we get,

\(\begin{aligned}{} - 4956.8 = - {f_k} \cdot \left( {82} \right)\\ \Rightarrow {f_k} = 1982.7\end{aligned}\)

Hence, the average force exerted on her by the snowdrift as it stops her is \({f_k} = 1982.7\)N.