A boy weighing 4.00 * 10^2 N jumps from a height of 2.00 m to the ground below.
ID: 2024865 • Letter: A
Question
A boy weighing 4.00 * 10^2 N jumps from a height of2.00 m to the ground below. Assume that the force of the
ground on his feet is constant.
(a) Compute the force of the ground on his feet if he
jumps stiff-legged, the ground compresses 2.00 cm,
and the compression of tissue and bones is negligible.
(b) Compute the force his legs exert on his upper body
(trunk, arms, and head), which weighs 2.50 * 10^2 N,
under the conditions assumed above.
(c) Now suppose that his knees bend on impact, so that
his trunk moves downward 40.0 cm during deceleration.
Compute the force his legs exert on his upper
body.
Explanation / Answer
(a) Given that the weight,w = 4.00 * 10^2 N ==> mass m = 4.00 * 10^2/9.8 = 40.816 kg the avg speed, v = 2gh = 2*9.8*2 = 6.26 m/s the momentum of the boy, P = mv = 255.510 kg m/s the change momentum of the boy = P-0 = 255.510 kg m/s-0 = 255.510 kg m/s Therefore the Impulse of ground against boy = change in momentum of boy = 255.510 N-sImpulse of ground = Force x time = 255.510 N-s
Depression of ground ,d = 2m-0.02 m = 1.98 m =>It is known the boy has an initial speed (as he strikes the ground) = 6.26 m/s and
he has a final speed = 0 m/s. And since his force against ground is constant,
his AVG speed during "ground stop" = (6.26 + 0)/2 = 3.13 m/s
Distance , d = (AVG speed) x (t)
1.98 = 3.13t
time, t = 1.98/3.13 = 0.632 s
Impulse = (F) x (t)
F = 255.510/t = 403.91 N the change momentum of the boy = P-0 = 255.510 kg m/s-0 = 255.510 kg m/s Therefore the Impulse of ground against boy = change in momentum of boy = 255.510 N-s
Impulse of ground = Force x time = 255.510 N-s
Depression of ground ,d = 2m-0.02 m = 1.98 m =>It is known the boy has an initial speed (as he strikes the ground) = 6.26 m/s and
he has a final speed = 0 m/s. And since his force against ground is constant,
his AVG speed during "ground stop" = (6.26 + 0)/2 = 3.13 m/s
Distance , d = (AVG speed) x (t)
1.98 = 3.13t
time, t = 1.98/3.13 = 0.632 s
Impulse = (F) x (t)
F = 255.510/t = 403.91 N (b) here w=2.5*10^2 N mass, m = 25.51kg the change in momentum, p = 159.69 kg m/s from the abve problem, F = impulse /time = 252.68 N (c) Repeat same as above, the distance d = 1.96 m from this , time t = 1.96/3.13 = 0.626 s Impulse = (F) x (t)
F = 159.69/0.626 = 255 N = 255 N
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