Question

A person slaps her leg with her hand, which results in her hand coming to rest in a time interval of 2.65 ms from an initial speed of 5.25 m/s . What is the magnitude of the average contact force exerted on the leg, assuming the total mass of the hand and the forearm to be 1.75 kg

Answer 1

Answer:

the magnitude of the average contact force exerted on the leg is  3466.98 N

Explanation:

Given the data in the question;

Initial velocity of hand v₀ = 5.25 m/s

final velocity of hand v = 0 m/s

time interval t = 2.65 ms = 0.00265 s

mass of hand m = 1.75 kg

We calculate force on the hand F$_{hand$

using equation for impulse in momentum

F$_{hand$ × t = m( v - v₀ )

we substitute

F$_{hand$ × 0.00265  = 1.75( 0 - 5.25 )

F$_{hand$ × 0.00265  = 1.75( - 5.25 )

F$_{hand$ × 0.00265  = -9.1875

F$_{hand$ = -9.1875 / 0.00265  

F$_{hand$ = -3466.98 N

Next we determine force on the leg F$_{leg$

Using Newton's third law of motion

for every action, there is an equal opposite reaction;

so, F$_{leg$ = - F$_{hand$

we substitute

F$_{leg$ = - ( -3466.98 N )

F$_{leg$ = 3466.98 N

Therefore, the magnitude of the average contact force exerted on the leg is  3466.98 N