A 77 kg sprinter is running the 100 m dash. At one instant, early in the race, h

Question

A 77 kg sprinter is running the 100 m dash. At one instant, early in the race, his acceleration is 4.7 m/s^2.

1. What total force does the track surface exert on the sprinter? Assume his acceleration is parallel to the ground. Give your answer as a magnitude.

2. What total force does the track surface exert on the sprinter? Assume his acceleration is parallel to the ground, and give your answer as an angle with respect to the horizontal.

3. This force is applied to one foot (the other foot is in the air), which for a fraction of a second is stationary with respect to the track surface. Because the foot is stationary, the net force on it must be zero. Thus the force of the lower leg bone on the foot is equal but opposite to the force of the track on the foot. If the lower leg bone is 60 degrees from horizontal,what is the components of the leg’s force on the foot in the direction parallel to the leg? (Force components perpendicular to the leg can cause dislocation of the ankle joint.)

Explanation / Answer

mass of the sprinter is m =77kg acceleration due to gravity of sprinter is g =9.8m/s^2 is works down words on sprinter the horizontal acceleration of the sprinter is a = 4.7m/s^2 1. the resultant acceleration of the sprinter is ar=a^2+g^2 =4.7^2+9.8^2                                                                   ar =118.13 =10.86m/s^2 the force exerted on sprinter by track surface is F = m ar                                                                           = (77kg)*(10.86m/s^2)                                                                           =836.89 N 2. if the acceleration of the sprinter is made an angle with horizontal i.e the horizontal acceleration is ah = a cos =4.7 cos the resultant acceleration of the sprinter is ar=(4.7 cos)^2+g^2 =(4.7 cos)^2+9.8^2                                                                   ar =(4.7 cos)^2+9.8^2 m/s^2 the force exerted on sprinter by track surface is F = m ar                                                                           = (77kg)*((4.7 cos)^2+9.8^2)N 3. If the lower leg bone is 600 from horizontal the horizontal acceleration is    ah = a cos60 =4.7 *0.5= 2.35m/s^2 the resultant acceleration of the sprinter is ar=(2.35)^2+(9.8)^2                                                                                                                                  ar =10.07 m/s^2 if one leg is in the air the force exerted on sprinter by track surface is F = m ar /2                                                                                                              =(77kg)*(10.07)/2                                                                                                             =387.99 N in this case the force exerted only one leg by this cause we take half force                                                                           = (77kg)*(10.86m/s^2)                                                                           =836.89 N 2. if the acceleration of the sprinter is made an angle with horizontal i.e the horizontal acceleration is ah = a cos =4.7 cos the resultant acceleration of the sprinter is ar=(4.7 cos)^2+g^2 =(4.7 cos)^2+9.8^2                                                                   ar =(4.7 cos)^2+9.8^2 m/s^2 the force exerted on sprinter by track surface is F = m ar                                                                           = (77kg)*((4.7 cos)^2+9.8^2)N 3. If the lower leg bone is 600 from horizontal the horizontal acceleration is    ah = a cos60 =4.7 *0.5= 2.35m/s^2 the resultant acceleration of the sprinter is ar=(2.35)^2+(9.8)^2                                                                                                                                  ar =10.07 m/s^2 if one leg is in the air the force exerted on sprinter by track surface is F = m ar /2                                                                                                              =(77kg)*(10.07)/2                                                                                                             =387.99 N in this case the force exerted only one leg by this cause we take half force the resultant acceleration of the sprinter is ar=(4.7 cos)^2+g^2 =(4.7 cos)^2+9.8^2                                                                   ar =(4.7 cos)^2+9.8^2 m/s^2 the force exerted on sprinter by track surface is F = m ar                                                                           = (77kg)*((4.7 cos)^2+9.8^2)N 3. If the lower leg bone is 600 from horizontal the horizontal acceleration is    ah = a cos60 =4.7 *0.5= 2.35m/s^2 the resultant acceleration of the sprinter is ar=(2.35)^2+(9.8)^2                                                                                                                                  ar =10.07 m/s^2 if one leg is in the air the force exerted on sprinter by track surface is F = m ar /2                                                                                                              =(77kg)*(10.07)/2                                                                                                             =387.99 N in this case the force exerted only one leg by this cause we take half force                                                                           = (77kg)*((4.7 cos)^2+9.8^2)N 3. If the lower leg bone is 600 from horizontal the horizontal acceleration is    ah = a cos60 =4.7 *0.5= 2.35m/s^2 the resultant acceleration of the sprinter is ar=(2.35)^2+(9.8)^2                                                                                                                                  ar =10.07 m/s^2    ah = a cos60 =4.7 *0.5= 2.35m/s^2 the resultant acceleration of the sprinter is ar=(2.35)^2+(9.8)^2                                                                                                                                  ar =10.07 m/s^2 the resultant acceleration of the sprinter is ar=(2.35)^2+(9.8)^2                                                                                                                                  ar =10.07 m/s^2 if one leg is in the air the force exerted on sprinter by track surface is F = m ar /2                                                                                                              =(77kg)*(10.07)/2                                                                                                             =387.99 N in this case the force exerted only one leg by this cause we take half force

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