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Mechanical energy is the sum of kinetic energy (energy of motion) and potential energy (energy stored in a system due to the location of its elements).

The change in the mechanical strength of the system equals the work done in the system by external force and gravitational force.

The change in mechanical energy of the block is equal to the work done on kit by the kinetic friction force.

\(\begin{aligned}{}{E_f} - {E_i} &= {W_{other}}\\\left( {{K_f} + {U_{g,f}}} \right) - \left( {{K_i} + {U_{g,i}}} \right) &= \overrightarrow {{f_k}} \cdot \overrightarrow d \end{aligned}\)

Chose the zero level to be at the bottom of the ramp, which means\({U_{g,f}}\)at that point is zero.

Now, the kinetic friction force acts in the opposite direction to the displacement \(d\).

So, the initial gravitational potential energy depends on the initial vertical height of the crate, which is as shown in figure below.

\(h = d\sin \left( {52} \right)\)

Now,

\(\begin{aligned}{}{E_f} - {E_i} &= {W_{other}}\\\left( {\frac{1}{2}m{v_f}^2 + 0} \right) - \left( {0 - mgh} \right) &= - {\mu _k}Nd\\\frac{1}{2}m{v_f}^2 + mgh &= - {\mu _k}N\left( {\frac{h}{{\sin 52^\circ }}} \right)\end{aligned}\)

Here, \(m\) is the mass, \({v_f}\) is the velocity, \(g\) is the acceleration due to gravity, \(N\) is the normal force, and \(h\) is the height.

Now, from the diagram you can write the normal force as,

\(N = mg\cos 52^\circ \)

\(\begin{aligned}{}\frac{1}{2}m{v^2} + mgh &= - {\mu _k}mg\cos 52^\circ \left( {\frac{h}{{\sin 52^\circ }}} \right)\\\frac{1}{2}{v^2} + gh &= - h{\mu _k}g\cot 52^\circ \\\frac{1}{2}{v^2} &= - \left( {gh + h{\mu _k}g\cot 52^\circ } \right)\\{v^2} &= - 2gh\left( {1 + {\mu _k}\cot 52^\circ } \right)\end{aligned}\)

\(h = \frac{{{v^2}}}{{2g\left( {1 - {\mu _k}\cot 52^\circ } \right)}}\) 鈥�.. (1)

Consider the known data as below.

The speed,\(v = 4{\rm{ }}{m \mathord{\left/ {\vphantom {m s}} \right.\\} s}\)

Acceleration due to gravity,\(g = 9.8{\rm{ }}{m \mathord{\left/ {\vphantom {m {{s^2}}}} \right.\\} {{s^2}}}\)

Now, for cast iron you have, the coefficient of sliding friction as,

\({\mu _k} = 0.15\)

Therefore, the height for the cast iron is,

\(\begin{aligned}{}{h_{ci}} &= \frac{{{{\left( 4 \right)}^2}}}{{2\left( {9.8} \right)\left( {1 - \left( {0.15 \times \cot 52^\circ } \right)} \right)}}\;\\ &= \frac{{16}}{{19.6\left( {1 - \left( {0.15 \times 0.78} \right)} \right)}}\\ &= \frac{{16}}{{19.6 \times 0.883}}\;\\ &= 0.92{\rm{ }}m\end{aligned}\)

Now, for copperyou have, the coefficient of sliding friction as,

\({\mu _k} = 0.29\)

Therefore, the height for the copper is,

\(\begin{aligned}{}{h_{cu}} &= \frac{{{{\left( 4 \right)}^2}}}{{2\left( {9.8} \right)\left( {1 - \left( {0.29 \times \cot 52^\circ } \right)} \right)}}\\ &= \frac{{16}}{{19.6\left( {1 - \left( {0.29 \times 0.78} \right)} \right)}}\end{aligned}\)

\(\begin{aligned}{}{h_{cu}} &= \frac{{16}}{{19.6 \times 0.773}}\;\;\\ &= 1.06{\rm{ }}m\end{aligned}\)

Now, for zincyou have, the coefficient of sliding friction as,

\({\mu _k} = 0.85\)

Therefore, the height for the zinc is,

\(\begin{aligned}{}{h_{zn}} &= \frac{{{{\left( 4 \right)}^2}}}{{2\left( {9.8} \right)\left( {1 - \left( {0.85 \times \cot 52^\circ } \right)} \right)}}\;\\ &= \frac{{16}}{{19.6\left( {1 - \left( {0.85 \times 0.78} \right)} \right)}}\end{aligned}\)

\(\begin{aligned}{}{h_{zn}} &= \frac{{16}}{{19.6\left( {1 - 0.663} \right)}}\\ &= \frac{{16}}{{19.6 \times 0.337}}\\ &= 2.42{\rm{ }}m\end{aligned}\)

Write the equation for height as below.

\(h = \frac{{{v^2}}}{{2g\left( {1 - {\mu _k}\cot 52^\circ } \right)}}\;\)

So, \(h\) does not depend on mass.

Hence, for the \(m = 0.84{\rm{ }}kg\), the height of the copper is \({h_{cu}} = 1.06{\rm{ }}m\).

Rearrange equation (1) for the velocity as below.

\(\begin{aligned}{}h &= \frac{{{v^2}}}{{2g\left( {1 - {\mu _k}\cot 52^\circ } \right)}}\\{v^2} &= 2gh\left( {1 - {\mu _k}\cot 52^\circ } \right)\end{aligned}\)

If \(h\) remains constant. Therefore,

\({v^2} \propto \left( {1 - {\mu _k}\cot 52^\circ } \right)\)

So, if \(\theta \) will increase, \(\cot \theta \) decrease.

Then the term \(\left( {1 - {\mu _k}\cot \theta } \right)\) will increase. Which means the velocity \(v\) also increase.

Hence when \(\theta \) is increase and \(h\) remains same, \(v\) increase.