Question:(II) A sled is initially given a shove up a frictionless 23.0° incline. It reaches a maximum vertical height 1.22 m higher than where it started at the bottom. What was its initial speed?

The sled is initially at the bottom of the inclined plane. It then moves upward toward the top of the inclined plane.

According to energy conservation, the total energy at the bottom of the inclined plane equals the total energy at the maximum height.

The energy at the bottom and the top is the summation of the kinetic and the potential energies.

At the top, some potential energy exists but the kinetic energy is zero. At the bottom, there is kinetic energy but the potential energy is zero.

The given data can be listed below as:

• The angle of inclination of the inclined plane is\(\theta = 25^\circ \).
• The acceleration due to gravity is\(g = 9.81{\rm{ m/}}{{\rm{s}}^2}\).
• The maximum height of the sled is \({h_f} = {\rm{1}}{\rm{.22 m}}\).

The diagram can be shown as:

Here,\({F_N}\)is the normal reaction force, g is the acceleration due to gravity, mgis the weight of the sled, m is the mass of the sled, and\(\theta \)is the inclination angle. The given surface is frictionless.

According to the conservation of energy, the energies can be written as:

\(\begin{array}{c}{E_i} = {E_f}\\P.{E_i} + K.{E_i} = P.{E_f} + K.{E_f}\\mg{h_i} + \frac{1}{2}mv_i^2 = mg{h_f} + \frac{1}{2}mv_f^2\end{array}\)

\(0 + \frac{1}{2}mv_i^2 = mg{h_f} + 0\) … (i)

Here,\({E_i}\)is the total energy at the bottom, and\({E_f}\)is the total energy at the top.

The initial height\({h_i}\)is equal to zero,\({h_f}\)is the final height at the top,\({v_i}\)is the initial velocity at the bottom, and\({v_f}\)is the final velocity at the top, which is equal to zero.

The initial potential energy \(P.{E_i}\) is equal to zero, \(K.{E_i}\) is the initial kinetic energy at the bottom, \(P.{E_f}\)is the final potential energy at the top, and \(K.{E_f}\) is the final kinetic energy at the top, which is equal to zero.

From equation (i), the initial velocity of the sled can be expressed as:

\(\begin{array}{c}\frac{1}{2}mv_i^2 = mg{h_f}\\v_i^2 = 2g{h_f}\end{array}\)

\({v_i} = \sqrt {2g{h_f}} \) … (ii)

Substitute the values in the above expression.

\(\begin{array}{c}{v_i} = \sqrt {2g{h_f}} \\ = \sqrt {2 \times 9.81{\rm{ m/}}{{\rm{s}}^2} \times {\rm{1}}{\rm{.22 m}}} \\ = 4.89{\rm{ m/s}}\end{array}\)

Thus, the initial velocity at the bottom is 4.89 m/s.