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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
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