Question

A parachutist weighs 1000N. when she opens her parachute, it pulls upwards on her with a force of 2000N. (a) draw a diagram to show forces acting on the parachutist. (b) calculate the resultant forces acting on the parachutist. (c) what effect will this force have on her ?​

Answer 1

(a) Find free-body diagram in attachment (please rotate the picture, such that R points upward and mg points downward)

There are only two forces acting on the parachutist:

- Its weight, downward, of magnitude 1000 N, labelled with "mg" in the diagram (where m = mass of the parachutist, g = acceleration of gravity)

- The air resistance, upward, of magnitude 2000 N, labelled with "R" in the diagram

As the air resistance is larger than the weight, in the diagram it is represented with a longer arrow, in order to show the difference in magnitude.

(b) 1000 N, upward

By taking upward as positive direction, we can rewrite the two forces as:

R = +2000 N

mg = -1000 N

Where we have written the weight as a negative number, since its direction is downward.

Therefore, the net force on the parachutist will be

F = R + mg = +2000 + (-1000) = +1000 N

And the positive sign indicates that the resultant force is upward.

(c) The parachutist will be accelerated upward (= he/she will slow down)

We can answer this part by applying Newton's second law, which is summarized by the following:

`F=ma` (1)

where

F is the resultant force on the body

m is its mass

a is its acceleration

For the parachutist in this problem, the mass is

`m=(mg)/(g)=(1000 N)/(9.8 N/kg)=102 kg`

Therefore, using (1), we find the acceleration of the parachutist:

`a=(F)/(m)=(+1000)/(102)=+9.8 m/s^2`

Where the positive sign indicates that the acceleration is upward. Therefore, the parachutist will be accelerated upward, which means that he/she will slow down, since its direction of motion was downward.