Gravitational Acceleration Problem

[BandGeek13]

Homework Statement

A 95 kg man jumps, therefore exerting a force of 1000 N on a bathroom scale. The scale is in a motionless elevator in Earth's gravity. What is the man's acceleration?

Homework Equations

Newton's 2nd Law: F=ma
Possibly Fg=mg ?

The Attempt at a Solution

I tried just using F=ma.
therefore, it'd be:
1000 N=(95kg)(a)
a=10 m/s^2

But this does not take into consideration the Earth's gravity.

 

[Redbelly98]

Welcome to PF.

"F=ma" uses the net force (vector sum of all forces acting on the man)

Your F uses the 1000N force that the scale exerts on the man. As you suspect, you need to add the force due to Earth's gravity. Just take the vector sum of the scale force and the force due to gravity.

 

[BandGeek13]

So would it be:

F(normal)-F(gravity)=ma
F(normal) - mg=ma
1000 N [up] - (95 kg)(9.8 N/kg [down])=(95 kg)(a)
1000 N [up] - 931 N [down] = (95 kg)(a)
69 N [up] = (95 kg)(a)
0.73 m/s^2 = a

This seems very low..

 

[BandGeek13]

wait!

instead of:
69 N [up] = (95 kg)(a)
0.73 m/s^2 = a

is it:
1931 N [up] = (95 kg)(a)
20. m/s^2 = a

 

[Redbelly98]

So would it be:
.
.
.
0.73 m/s^2 = a

This seems very low..

Looks good to me.

Note, the man's weight is about 930 N, so the scale would read 930 N when a=0.

The scale reading of 1000N is not much more than this, so expect a to be considerably less than g=9.8 m/s^2.

An acceleration of g=9.8 m/s^2 upward would require a scale reading of 2*930N, which would give F_net = +930 N upward.

EDIT:

wait!
.
.
.
is it:
1931 N [up] = (95 kg)(a)
20. m/s^2 = a

Nope. We have +1000N (upward) and -931N (downward), for F_net=31N upward

 

[BandGeek13]

Alright.
Thank you!