The feet of a standing person of mass m exert a force equal to mg on the floor, and the floor exerts an equal and opposite force upwards on the feet, which we call the normal force. During the extension phase of a vertical jump (see page 145), the feet exert a force on the floor that is greater than mg, so the normal force is greater than mg. As you learned in Chapter 4, we can use this result and Newton’s second law to calculate the acceleration of the jumper: a = Fnet/m = (n - mg)/m.Using energy ideas, we know that work is performed on the jumper to give him or her kinetic energy. But the normal force can’t perform any work here, because the feet don’t undergo any displacement. How is energy transferred to the jumper?
The feet of a standing person of mass m exert a force equal to mg on the floor, and the floor exerts an equal and opposite force upwards on the feet, which we call the normal force. During the extension phase of a vertical jump (see page 145), the feet exert a force on the floor that is greater than mg, so the normal force is greater than mg. As you learned in Chapter 4, we can use this result and Newton’s second law to calculate the acceleration of the jumper: a = Fnet/m = (n - mg)/m.Using energy ideas, we know that work is performed on the jumper to give him or her kinetic energy. But the normal force can’t perform any work here, because the feet don’t undergo any displacement. How is energy transferred to the jumper?
Physics for Scientists and Engineers: Foundations and Connections
1st Edition
ISBN:9781133939146
Author:Katz, Debora M.
Publisher:Katz, Debora M.
Chapter5: Newton's Laws Of Motion
Section: Chapter Questions
Problem 39PQ: A student takes the elevator up to the fourth floor to see her favorite physics instructor. She...
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The feet of a standing person of mass m exert a force equal to mg on the floor, and the floor exerts an equal and opposite force upwards on the feet, which we call the normal force. During the extension phase of a vertical jump (see page 145), the feet exert a force on the floor that is greater than mg, so the normal force is greater than mg. As you learned in Chapter 4, we can use this result and Newton’s second law to calculate the acceleration of the jumper: a = Fnet/m = (n - mg)/m.
Using energy ideas, we know that work is performed on the jumper to give him or her kinetic energy. But the normal force can’t perform any work here, because the feet don’t undergo any displacement. How is energy transferred to the jumper?
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