A crate of mass $\mathit{M}$starts from rest at the top of a friction less ramp inclined at an angle $\mathit{\alpha }$ above the horizontal. Find its speed at the bottom of the ramp, a distance $\mathit{d}$ from where it started. Do this in two ways: Take the level at which the potential energy is zero to be (a) at the bottom of the ramp with $\mathit{y}$positive upward, and (b) at the top of the ramp with $\mathit{y}$positive upward. (c) Why didn’t the normal force enter into your solution?

The given data can be listed below as,

• The mass of a crate is, $M$.
• The initial speed of the crate at the top of the ramp is, $u=0$(Since start from rest)
• The angle of inclination of the ramp with the horizontal is, $\alpha$.
• The distance between top and bottom of the ramp is, $d$.

In this question, the concept of energy conservation helps to determine the speed of the crate along the inclined path of the ramp under the consideration of the different reference points. The value of the gravitational potential energy depends on the consideration of the reference point.

The free-body diagram of the crate is expressed below.

Here, $N$ is the normal reaction force on the crate by the ramp and $W$is the weight of the crate.

According to energy conservation method, the relation of speed of the crate at the bottom of the ramp when the potential energy is zero at the bottom of the ramp with positive upward is expressed as,

$$\begin{gathered} E_i=E_f \\ \left[{\left(KE\right)}_i+{\left(PE\right)}_i\right]=\left[{\left(KE\right)}_f+{\left(PE\right)}_f\right] \\ \left[\frac{1}{2}M{u}^2+Mgy\right]=\left[\frac{1}{2}M{v}^2+{\left(PE\right)}_f\right].....\left(1\right) \end{gathered}$$

Here, $E_i$is the total energy of the crate at the top of the ramp, $E_i$ is the total energy of the crate at the bottom of the ramp, ${\left(KE\right)}_i$ is the kinetic energy of the crate at the top of the ramp, ${\left(PE\right)}_i$ is the gravitational potential energy at the top of the ramp, ${\left(KE\right)}_i$is the kinetic energy of the crate at the bottom of the ramp, ${\left(PE\right)}_i$is the gravitational potential energy at the bottom of the ramp whose value is equal to zero means ${\left(PE\right)}_f=0$, $v$is the speed of the crate at the bottom of the ramp, $y$is the vertical height of the top of the ramp whose value can be expressed from the free-body diagram as, $y=d\sin \alpha$ and $g$ is the gravitational acceleration whose value is$9.81m/s^2$.

Substitute all the known values in the equation (1).

$$\begin{gathered} \left[\frac{1}{2}M{\left(0\right)}^2+Mg\left(d\sin \alpha \right)\right]=\left[\frac{1}{2}M{v}^2+0\right] \\ Mgd\sin \alpha =\frac{1}{2}M{v}^2 \\ gd\sin \alpha =\frac{1}{2}{v}^2 \\ v=\sqrt{2gd\sin \alpha }.....\left(2\right) \end{gathered}$$

Substitute all the known values in the equation (2).

$$\begin{gathered} v=\sqrt{2\left(9.81{\text{m/s}}^{\text{2}}\right)d\sin \alpha } \\ \approx 4.43\sqrt{d\sin \alpha } \end{gathered}$$

Thus, the speed of the crate at the bottom of the ramp when the potential energy is zero at the bottom of the ramp is$4.43\sqrt{d\sin \alpha }$.

According to energy conservation method, the relation of speed of the crate at the bottom of the ramp when the potential energy is zero at the top of the ramp with positive upward is expressed as,

$$\begin{gathered} E_i=E_f \\ \left[{\left(KE\right)}_i+{\left(PE\right)}_i\right]=\left[{\left(KE\right)}_f+{\left(PE\right)}_f\right] \\ \left[\frac{1}{2}M{u}^2+{\left(PE\right)}_i\right]=\left[\frac{1}{2}M{v}^2+\left(-Mgy\right)\right] \\ \left[\frac{1}{2}M{u}^2+{\left(PE\right)}_i\right]=\left[\frac{1}{2}M{v}^2-\left(Mgy\right)\right].....\left(3\right) \end{gathered}$$

Here, $E_i$is the total energy of the crate at the top of the ramp, $E_f$is the total energy of the crate at the bottom of the ramp, ${\left(KE\right)}_i$is the kinetic energy of the crate at the top of the ramp, ${\left(PE\right)}_i$is the gravitational potential energy at the bottom of the ramp whose value is equal to zero means ${\left(PE\right)}_i=0$, ${\left(KE\right)}_f$is the kinetic energy of the crate at the bottom of the ramp, $v$is the speed of the crate at the bottom of the ramp, ${\left(PE\right)}_f$is the gravitational potential energy at the bottom of the ramp, $y$is the vertical height of the top of the ramp whose value can be expressed from the free-body diagram as, $y=d\sin \alpha$.$y=d\sin \alpha$.

Substitute all the known values in the equation (3).

$$\begin{gathered} \left[\frac{1}{2}M{\left(0\right)}^2+0\right]=\left[\frac{1}{2}M{v}^2-Mgd\sin \alpha \right] \\ 0=\left[\frac{1}{2}M{v}^2-Mgd\sin \alpha \right] \\ gd\sin \alpha =\frac{1}{2}{v}^2 \\ v=\sqrt{2gd\sin \alpha }.....\left(4\right) \end{gathered}$$

Substitute all the known values in the equation (4).

$$\begin{gathered} v=\sqrt{2\left(9.81{\text{m/s}}^{\text{2}}\right)d\sin \alpha } \\ \approx 4.43\sqrt{d\sin \alpha } \end{gathered}$$

Thus, the speed of the crate at the bottom of the ramp when the potential energy is zero at the top of the ramp is$4.43\sqrt{d\sin \alpha }$.

In the above calculations, the speed of the crate at the bottom of the ramp under two different conditions does not include normal reaction force because the motion of the crate takes place only along the inclined ramp in the downward direction, whereas no motion takes place along the normal reaction force (perpendicular to the inclined plane of the ramp). The required forces in the calculation of the speed of the crate at the bottom of the ramp are only the force involved along the inclined path of the ramp.

Thus, the normal force didn’t enter in the solution of the speed of the crate at the bottom of the crate.