Tension of a rope holding a platform and a person

[Elphaba]

An 80 kg person standing on a 20 kg plateform is suspended by a rope passing over a stationary pulley that is free to rotate. The other end of the rope is held by the person. The masses of the rope and pulley are negligible. You may use g=10 m/s/s. Assume that friction is negligible and the parts of the rope shown remain verticle.


ok.. so the whole system is 100kg = 1000N right?

(a) if the platform and the person are at rest, what is the tension in the rope?
T=NetForce-Mg (i think)

so mg = 1000N
and NF= 1000N too??


The person now pulls on the rope so that the acceleration of the person and the platform is 2m/s/s upward.

(b) What is the tension in the rope under these new conditions?
would NF change?

(c) Under these new conditions, what is the force exerted by the platform on the person?

F= ma
so... platform on person... 20kg on 80 kg.. accel is 2...
um... confused...what do i do now?

After a short time, the person and the platform reach and sustain and upward velocity of 0.4m/s

(d) Determine the power output of the person required to sustain the velocity.


Hm... he hasn't taught us power yet... A little help please?

 

[primarygun]

$$Power=Energy/Time$$

 

[primarygun]

For (a), There are one rope which is divided into two parts.
Both are vertical, one end is the lift, one end is the pulley;;;;;;
One end is the man, the other is the pulley.
For (b), if you can understand and get the correct answer of (a). That cannot trick you.
For c, Just need to think of the lift and the man.
F=ma---->Upward forces - weights= ma I arranged the terms since I know which force is bigger.
For (4), to find the energy, it's better for you to find out in 1 second, the change in PE and KE.
Then you can get the power.

 

[Elphaba]

?
ok...
yep
but what is the Tension?
100kg?
upward force = 1000N... Weights is 100kg?? = 100*2?
um... i don't get the last 2 sentences.