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 vâ‚€ = 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[tex]_{hand[/tex]
using equation for impulse in momentum
F[tex]_{hand[/tex] × t = m( v - v₀ )
we substitute
F[tex]_{hand[/tex] × 0.00265  = 1.75( 0 - 5.25 )
F[tex]_{hand[/tex] × 0.00265  = 1.75( - 5.25 )
F[tex]_{hand[/tex] × 0.00265  = -9.1875
F[tex]_{hand[/tex] = -9.1875 / 0.00265 Â
F[tex]_{hand[/tex] = -3466.98 N
Next we determine force on the leg F[tex]_{leg[/tex]
Using Newton's third law of motion
for every action, there is an equal opposite reaction;
so, F[tex]_{leg[/tex] = - F[tex]_{hand[/tex]
we substitute
F[tex]_{leg[/tex] = - ( -3466.98 N )
F[tex]_{leg[/tex] = 3466.98 N
Therefore, the magnitude of the average contact force exerted on the leg is  3466.98 N
