Frictionless pulley Forces Problem

AI Thread Summary
An 85kg man lowers himself from a height of 10.0m using a rope over a frictionless pulley connected to a 65kg sandbag. The initial calculations for the tension force (F_T) were incorrect, initially reported as 58.93N but later corrected to 721.93N. This adjustment led to a recalculated acceleration of 1.30m/s². Using this acceleration, the final velocity of the man upon hitting the ground was determined to be 5.11m/s. The problem-solving process highlighted the importance of accurately calculating forces in physics problems.
ubiquinone
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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
 
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You'll need to check your working for F_T. It's about 700+ N, not 58.93 N.
 
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
 
yep that's it!
 
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