Question

Very large forces are produced in joints when a person jumps from some height to the ground. Calculate the force produced if an 79.0–kg person jumps from a 0.510–m-high ledge and lands stiffly, compressing joint material 1.30 cm as a result. (Be certain to include the weight of the person.) 3.04*10^4J Incorrect. Tries 1/7 Previous Tries In practice the knees bend almost involuntarily to help extend the distance over which you stop. Calculate the force produced if the stopping distance is 0.345 m. 1.92×103 J You are correct. Find the ratio of the force in part (a) with the weight of the person. Tries 0/7 Find the ratio of the force in part (b) with the weight of the person.

Answer 1

Answer:

A) 31 kJ

B)  1.92 KJ

C) 40 ,  2.48

Explanation:

weight of person ( m ) = 79 kg

height of jump ( h ) = 0.510 m

Compression of joint material ( d ) = 1.30 cm ≈  0.013 m

A) calculate the force

Fd = mgh

F = mgh / d

W = mg

F(net) = W + F  = mg ( 1 + $\frac{h}{d} )$

          =  79 * 9.81 ( 1 + (0.51 / 0.013) )

          = 774.99 ( 40.231 ) ≈ 31 KJ

B) calculate the force when the stopping distance = 0.345 m

d = 0.345 m

Fd = mgh  hence  F = mgh / d

F(net) = W + F  = mg ( 1 + $\frac{h}{d} )$

          =  79 * 9.81 ( 1 + (0.51 / 0.345) )

          = 774.99 ( 2.478 ) = 1.92 KJ

C) Ratio of force in part a with weight of person

=  31000 / ( 79 * 9.81 ) = 31000 / 774.99 = 40

  Ratio of force in part b with weight of person

= 1920 / 774.99 = 2.48