While dangling a hairdryer by its cord, you observe that the cord is vertical when the hairdryer isoff and, once it is turned on, the hairdryer moves to the right and comes to rest when the cordmakes an angle of 5 degrees with the vertical, as shown below. In a different experiment, you determine that the same hairdryer is pushing 0.06 m3 of air through itself every two seconds. The mass ofthe hairdryer is 420 g. Determine the speed of the air leaving the hairdryer, v air. Assume that themass of 1 m3 air is 1.2 kg and that the hairdryer is blowing air perpendicular to the wire.

The speed of the air leaving the hairdryer ( ) = 10 m/s

Given data :

m = 0.420 kg

g = 9.8 m/s

∅ = 5°

v ( speed of air leaving the hairdryer ) = ?

First step ; determine the thrust value using the relation below

Thrust ( τ ) = Mg ( tan ∅ )

                  = 0.420 * 9.8 * ( tan 5° ) = 0.3601 N

2nd step : Express the relationship between the thrust value and air ejection speed

τ  =  ------ ( 1 )

= ?

= 0.03 m³ / s =  0.036 kg/s

τ  = 0.3601 N

Final step ; Calculate the value of the speed of air leaving the hair dryer

Insert values into equation  ( 1 )

∴ = = 10 m/s

Hence we can conclude that the speed of the air leaving the hairdryer is 10m/s.

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The speed of the air leaving the hairdryer is 10m/s.

Explanation:

The thrust that the dryer produces is what keeps it elevated at an angle of 5° from the vertical; therefore, from the force diagram we get

putting in , and solving for we get:

Now, this thrust produced is related to to the air ejection speed by the relation

where is the rate of air ejection which we know is

and since ,

putting this into equation (2) and the value of we get:

which gives

which is the speed of the air ejected.