A block of ice with mass 2.50 kg slides 2.00 m down an inclined plane that slopes downward at an angle of 36.9° below the horizontal. If the block of ice starts from rest, what is its final speed? Ignore friction.

The final speed is 4.85 m/s.

Explanation:

Given:

Mass of ice (m) = 2.50 kg

Length of incline (L) = 2.00 m

Angle of incline (x) = 36.9°

Initial speed of ice block (u) = 0 m/s

Final speed of ice block (v) = ?

Friction is ignored.

So, when friction is ignored, then, the change in kinetic energy equals change in potential energy. This is true from conservation of energy.

∴ Δ KE = Δ PE

Here, 'h' is the initial height of the block which can be determined from the right angled triangle as shown below.

From triangle ABC, AC = 2.00 m, ∠ACB = 36.9°, AB = 'h'

So, using sine of angle ratio, we have

Therefore, the final speed is given as:

Therefore, the final speed is 4.85 m/s.

force required to detach it is given by

charges are taken to be fundamental charge

now we have

so above is the minimum distance at which it will detach