Question

Calculate the absolute value of the magnitude of the impact force on a 60 kg parachutist falling at 10 m/s if the time to stop was 0.7 seconds.

Answer 1

Answer:

85.7 N

Explanation:

According to Newton's second law of motion, F is equivalent to the product of mass and acceleration, ma where m is the mass of parachute in kilograms and a is acceleration.

Acceleration is defined as the rate of change of velocity per unit time hence expressed as

$a=\frac {\triangle v}{t}=\frac {v-u}{t}$

Where v and u represent the final and initial velocities respectively.

Substituting this into initial equation of f then

$F=m\frac {v-u}{t}$

For this case, the final velocity is zero since it comes ro rest. Substituting 0 m/s dor v and 10 m/s for u then 0.7 s for t we can deduce that

$F=60\times \frac {0-10}{0.7}=−85.71428571428\approx -85.7N$