A 10 kg crate travels along a smooth slope that plots as y

A 10 kg crate travels along a smooth slope that plots as y = 1/4 x^3/2. At a point where x = 16 m its speed is 20 m/s. Determine the speed of the crate when it gets to the bottom of the slope (x = 0, y = 0).

Solution

We have a slope defined as y = 0.25x^3/2

Now, clearly the y component of the above equation denotes the distance travelled along the vertical direction whereas the x component is the distance travelled along the x direction. We know the speed of the crate at a point where x =16 m.

Using the equation of the slope, we will determine the vertical height for x = 16m and then conserve the total energy of the crate to determine the velocity it achieves by the time it reaches the bottom of the slope. Also, we will assume the level y = 0 as the reference level for calculating the potential energy.

Now for x = 16, we have: y= 0.25 x 4^3 = 16 m

So at the height of 16 metres the crate has a speed of 20 m/s, hence we equate the initial and final energy to get:

0.5 x 10 x 400 + 10 x 16 x 9.81 = 0.5 x 10 x v^2

or, v^2 = 32 x 9.81 + 400 = 713.92

or V = 26.7193 m/s

Therefore the required speed at the bottom of the slope is 26.7193 m/s