What Forces Act on the Body During a Chin-Up?

[Lavalamp22]

Homework Statement

People who do chin-ups raise their chin just over a bar (the chinning bar), supporting themselves only by their arms. Typically, the body below the arms is raised by about 30cm in a time of 1.0s , starting from rest. Assume that the entire body of a 760N person who is chinning is raised this distance and that half the 1.0s is spent accelerating and the other half decelerating, uniformly in both cases.

Homework Equations

S=volt+(1/2)at^2

The Attempt at a Solution

I thought I had the right answer, but mastering physics is saying it's wrong, and I don't know what's wrong.

.3=(0*.5)+.5*a*.5^2
.3=1/2a*.25
2.4=a
F=ma
F=(77.55)(2.4)
F=186.12N

 

[TheAbsoluTurk]

Seems like an energy-based approach may be in order.

 

[CWatters]

Why have you used S = 0.3m ?

As I read the question the body starts at the bottom, accelerates for 0.15m to some velocity V then decelerates for 0.15m becoming stationary at the top. It says assume uniform acceleration so I think you can use standard equations of motion.

In addition the force required will obviously be greater than 760N. You have forgotton something else.

 

[Lavalamp22]

Why have you used S = 0.3m ?

As I read the question the body starts at the bottom, accelerates for 0.15m to some velocity V then decelerates for 0.15m becoming stationary at the top. It says assume uniform acceleration so I think you can use standard equations of motion.

In addition the force required will obviously be greater than 760N. You have forgotton something else.

The question's wording is confusing me, but I understand what you mean. So it would be:

.15=(0*.5)+1/2a(.5^2)
a=1.2
F=(77.55)(1.2)
F=93.06N+760N=853.06N?

I only have 1 more attempt to answer the question before it will put a 0 as the grade, so please let me know if this is the correct answer. Thanks.