A 75.0 kg person steps on a scale in an elevator. The scale reads 761 N. What is the magnitude of the acceleration of the elevator?

Question

stormfire1 · Answer

The scale will read the total forces acting down on it (the sum of the normal forces). The forces pushing up on the person will be mg + m(elevator's acceleration) which will be opposed by a normal force of the same magnitude. So the equation would be:
$$Ftotal=761N=mg+Fn=(75.0kg∗9.8m/s2)+(75kg∗a)$$Now just solve for a.

stormfire1 · Answer

Also, another way to look at it is: $$\frac{761N}{75.0kg}= 10.1m/s^2$$That would be the elevator's acceleration if there were no gravitational forces acting on it.  So subtract the force of gravity from that to get:$$10.1m/s^2-9.8m/s^2 = 0.34 m/s^2 (upward)$$Either way you get the same answer.

anonymous · Answer

the computer is not accepting it as my answer

stormfire1 · Answer

I really can't help you there...I'm not familiar with how you're supposed to input it.  It may accept 10.1 m/s^2 but that's really not correct.  It should be either 0.3 m/s^2 or 0.34 m/s^2...depending on how you do the calculations

stormfire1 · Answer

I rounded my calcs based on significant figures...you could try doing them again without any rounding

anonymous · Answer

ok it accepted 0.333m/s^2 as the answer,i took g=9.81

anonymous · Answer

but why shuld we subtract from 9.81? cos if i were to take this in exam,i won't knw am to do dat

stormfire1 · Answer

You have to subtract 9.81 because the force of gravity is also pulling down on the scale