# Forces Problem

1. Oct 14, 2006

### ubiquinone

Hi there, I'm new to this forum but I'm very interested in maths and physics. I look forward to learning from the people here and hear some of your insights. Today I have a question involving forces. I've finished the problem and I was just wondering if anyone here may please check if it is correct. Thanks!!

Question: An $$85kg$$ man lowers himself to the ground from a height of $$10.0m$$ by holding onto a rope that runs over a frictionless pulley to a $$65kg$$ sandbag. With what speed does the man hit the ground if he started from rest?
If we begin by looking at the sac and the person separately we could set up two equations that describe the net force that is acting on them.
For the sac: $$\displaystyle F_{net_{sac}}=F_T-65g=65a$$
For the body: $$\displaystyle F_{net_{body}}=F_T-85g=85a$$
Since the magnitude of the acceleration, $$a$$ is equal for the sac and the body but opposite in direction, we can say,
$$\displaystyle\frac{F_T-65g}{65}=-\left(\frac{F_T-85g}{85}\right)$$
Solving, $$F_T=58.93N$$
Therefore, $$\displaystyle a=-\left(\frac{58.93-85g}{85}\right)=9.106m/s^2$$
To find the man's final velocity, we can use the formula $$v^2=v_1^2+2ad$$ where $$v_1=0.0m/s$$, $$a=9.106m/s^2$$ and $$d=10.0m$$
Thus, $$v_2=\sqrt{2(9.106m/s^2)(10.0m)}=13.5m/s$$

2. Oct 14, 2006

### Fermat

You'll need to check your working for F_T. It's about 700+ N, not 58.93 N.

3. Oct 14, 2006

### ubiquinone

Hi, thanks Fermat for replying. Yes, I made a calculation errror. $$F_T=721.93N$$. Therefore $$a=1.30m/s^2$$.
Substituting into the formula $$v_2=\sqrt{2(1.30m/s^2)(10.0m)}=5.11m/s$$

4. Oct 14, 2006

### Fermat

yep that's it!