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At the instant a race began, a 61-kg sprinter exerted a force of 730N on the starting block at a 22° angle with respect to the ground. What was the horizontal acceleration of the sprinter? If the force was exerted for 0.33s, with what speed did the sprinter leave the starting block?

Answer :

Answer

given,

mass of the sprinter = 61 kg

angle of the block = 22°

Force = 730 N

time = 0.33 s

Horizontal force = F cos θ

                           = 740 cos 22°

                           = 686.17 N

horizontal acceleration = [tex]\dfrac{F}{m}[/tex]

                                      =[tex]\dfrac{686.17}{61}[/tex]

                                      =11.25 m/s²

Impulse = F Δt

             = 686.17 x 0.33

            = 226.44 Kgm/s

velocity= [tex]\dfrac{I}{m}[/tex]

            =[tex]\dfrac{226.44}{61}[/tex]

            =3.71 m/s

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