Question

PART ONE:

A gust of wind blows an apple from a tree.
As the apple falls, the force of gravity on the
apple is 9.39 N downward, and the force of
the wind on the apple is 1.50 N to the right.
What is the magnitude of the net external
force on the apple?
Answer in units of N.

PART TWO:
What is the direction of the net external force
on the apple (measured from the downward
vertical, so that the angle to the right of
downward is positive)?
Answer in units of ◦

Answer 1

Part ONE: 9.51 N

PART TWO:
`9.08^(\circ)`

Step-by-step explanation:

Given information

Gravitational force exerted on the apple is 9.39 N

Force exerted by the wind on the apple is 1.5 N

Net force on apple is given by’

`R=\sqrt {F^(2)+W^(2)+2FWcos\theta}` where W is gravitational force, F is force due to wind and
`\theta` is the angle between the gravitational force and wind force. For this case,
`\theta=90^(\circ)` hence the equation is re-written as

`R=\sqrt {F^(2)+W^(2)}` since
`cos 90^(\circ)=0`

Then substituting W for 9.39 and F for 1.5 we obtain

`R=\sqrt{1.5^(2)+9.39^(2)}= 9.509054\approx 9.51 N`

The direction of net force is given by

`\phi=tan^(-1)(\frac {F}{W})`

`\phi=tan^(-1)(\frac {1.5}{9.39})=9.075998^(\circ)\approx 9.08^(\circ)`