Question

The figure is a graph of the force exerted by the floor on a woman making a vertical jump.

At what speed does she leave the ground? Hint: The force of the floor is not the only force acting on the woman.

Answer 1

(0.5mv^2 = w = )area under that graph
mg = 600
(0.5mv^2 = 600(0.4) + 0.2(1800)/2 + 0.1(2400)/2;;; v= 4.2m/s)

Answer 2

by impulse-momentum theorem,

F*dt = m*dv

m = 60

(0.4*600)+(0.1*1800)+(0.05*2400) = 60*dv

dv = 9 m/s

as initial velocity is 0,

final velocity is 9 m/s

Answer 3

F*dt/m
=((600+2400)/2 *0.2+0.5*2400*0.1)*0.3/60
=2.1 m/s

Answer 4

sol:
600 N of force = woman's weight.
At its peak,
force exerted on the woman by the floor is 2400 N,
so force is applied when jumping=2400 - 600
= 1800 N .

from diagram,
we see that the woman is in the air after 0.5 - 0.2
= 0.3 seconds.
Using the impulse momentum theorem:
we get
F?t = p = mv ==>
v = F?t/m = [(1800)(0.3)]/(600/9.8)
= 8.82 m/s.
as weight of the woman is 600 N, so her mass is 600/9.8
= 61.2 kg.) ans

Answer 5

3.8

Answer 6

as the woman pushes harder to make the jump the normal reaction increases initially and then as she looses contact with the ground reducs to zero

initial momentum of the woman = M(0)

final momentum = M(V)

the initial sraight line with 0 slope at 600N is th enormal reaction when thwe woman is just standing on the ground still.

∴ M*g = 600 ==> M = 61.22 kg

change in momentum = impulse by net force acting on the woman =

to find net force subtract a constant Mg due to gravity acting downwards apart from the force due to ground reaction

therefore in the graph make the zero line for net force as the horizontal through 600 N as shown below:

∴ V = 247.5 / 61.22 = 4.04 m/s

Answer 7

3.8

Answer 8

it's ACTUALLY 3.9 m/s This is definitely the right ans