A) show that the velocity ai the bottom of the slope is given by v opposes square root of H?

Question

and that the proportionality constant is

squareroot{2[g - f/msinO)

Answer 1

m = mass

O = angle of slope to horizontal

g = acceleration due to gravity

F = force of friction

a = acceleration

v = speed at bottom of slope

s = length of slope

h = vertical height.

Resolving parallel to the slope:

mg sin(O) - F = ma

g sin(O) - F / m = a

v^2 = 2as

= 2ah / sin(O)

Eliminating a:

v^2 / h = 2[ g - F / ( m sin(O) ) ].