Falling Objects & Terminal VelocityNotes p.24

IMPORTANT EXPLANATION of TERMINAL VELOCITYAs a falling object speeds up, the _____ _________ acting upwards ________ . This causes the unbalanced force and acceleration to ________ . Eventually the object will reach a speed where the upwards force due to ___ ________ balances the _______ of the object. From this point, the object will maintain a ________ ______ . We call this ________ ________ .

Although the acceleration due to gravity is 9.8 ms-2, downwards, objects do not maintain this acceleration in air because ________ ________________ acts upwards.air resistance

air resistanceincreases

decrease

air resistance weightconstant

velocity velocityterminal

Velo

city

(m

s-1)

Time (s)0

ExampleSketch a velocity – time graph to illustrate the motion of a skydiver from the moment she jumps from the plane. Your graph should clearly indicate the point at which she opens her parachute.

initially ‘a’ = 9.8 ms-2

‘a’ decreases as air resistance increases with speed.

‘a’ = 0 … balanced forcesTerminal Velocity.

Parachute opens

Velocity decreases as air resistance bigger than weight.But air resistance decreases with speed, so rate of deceleration decreases.

New “terminal velocity”when forces balance again.

Conservation of Energy

E p = mgh

Ek = ½ mv2

EW = Fd

From Nat 5 you should recall the following key energy equations:

Revision questions for Higher Physics, page 26, Q. 1 – 14.

E = P t

Notes p.25

Ep = m =g =

h =Ek =v =

Ew =F =d =E =P =t =

potential energy (in J)mass (in kg)

gravitational field strength (in Nkg-1)

height (in m)kinetic energy (in J)

velocity (in ms-1)

work done (in J)force (in N)distance (in m)energy transferred (in J)power (in W)

time (in s)

Ek = ½ mv2

d = EW / F

Worked ExampleA 20 g dart is travelling at 6 ms-1 when it strikes a dart board. The dartboard exerts an average frictional force of 30 N on the dart.Determine the dartboard thickness required to bring the dart to rest.

1st

2nd Kinetic energy changes to work done against friction!

Ek = ?m = 0.02 kgv = 6 ms-1

= 0.5 x 0.02 x 62 = 0.36 J

Ew = 0.36 JF = 30 Nd = ?

= 0.36 / 30 = 0.012 m = 1.2 cm

Revision questions for Higher Physics, page 26, Q. 1 – 14.