Question 18
A stone of mass m falls
from rest at the top of a cliff of height h into the sea below. Just before hitting the sea
the stone has speed v.
What is the average force of air resistance
acting on the stone during its fall?
Reference: Past Exam Paper – June 2017 Paper 12 Q16
Solution:
Answer:
C.
The stone possesses
gravitational potential energy at the top of the cliff.
As the stone falls, the
gravitational potential energy possesses by the stone is converted into kinetic
energy. However, some energy is lost due to work against air resistance.
From the conservation of energy,
GPE of stone at top = KE
of stone at bottom + Work done against air resistance
GPE of stone = mgh
KE of stone = ½ mv2
Work done against air
resistance = Resistive force × distance
moved against the force
Work done against air
resistance = F × h
GPE of stone at top = KE
of stone at bottom + Work done against air resistance
mgh = ½ mv2 +
Fh
F×h = mgh – ½ mv2
F×h = m (gh – ½ v2)
F = m (gh – ½ v2)
/ h
F = m (g – v2/2h)
where did the h go ? the h with gh before the final answer
ReplyDeletewhere did the h go with the gh before the final anser
ReplyDeleteboth terms in bracket are divided by h.
Deletegh / h = g
1/2 v^2 / h = v^2 / 2h