Question

# The diagram shows the normal force on Christine's feet vs. time, as recorded by a force...

The diagram shows the normal force on Christine's feet vs. time, as recorded by a force plate while she stands still initially (until point B), then jumps off the plate. (The graph is over-simplified and idealised, compared to reality.) When her feet leave the plate, the normal force is zero.
What is Christine's mass?
What is the magnitude of the (upward) impulse generated by the normal force of Christine during the time interval of her jump off the plate?
What is the magnitude of the downward impulse due to gravity during this interval?
What is the net impulse which propels her upwards when she jumps off the plate? (Recall, the net force on her is the normal force minus the force of gravity.)
What is her change in speed upwards for this process?

Normal Force (N) is the y-axis and Time (s) is the x-axis.

(a) Christine's mass which is given as :

According to graph, we have

the normal force acting on Christine static at 500 N until 1.5 sec have passed.

using an equation, FN = m g                                                                             { eq.1 }

(500 N) = m (9.8 m/s2)

m = 51.02 kg

(b) Magnitude of the (upward) impulse generated by the normal force of Christine during the time interval of her jump off the plate which is given as :

I = F t                                                                         { eq.2 }

where, F = normal force = (1500 - 500) N

t = time taken = (1.9 - 1.5) sec

inserting the values in eq.2,

I = (1000 N) (0.4 sec)

I = 400 N.s

(c) Magnitude of the downward impulse due to gravity during this interval which is given as :

using eq.2, I = F t

where, F = normal force = 1500 N

t = time taken = (2.1 - 1.9) sec

inserting the values in eq.2,

I = (1500 N) ( sec)

I = 300 N.s

(d) The net impulse which propels her upwards when she jumps off the plate which is given as :

Inet = (400 N.s)2 + (300 N.s)2

Inet = (160000 + 90000) N2.s2

Inet = 250000 N2.s2

Inet = 500 N.s

(e) Her change in speed upwards for this process which will be given as :

I = F t = m v

I = m v                                                                                   { eq.3 }

inserting the values in eq.3,

(400 N.s) = (51.02 kg) v

v = (400 N.s) / (51.02 kg)

v = 7.84 m/s