PHY-133 Section 60 Analytical Physies 1 3) Sandy is trainingto spend time on the

Question

PHY-133 Section 60 Analytical Physies 1 3) Sandy is trainingto spend time on the International Space Station by learning to wsik®, 3) platform Bloating in a pool of oil. The platform han a mass of 110. kg, is 5.00 m long, snd Roets consider "the system" to consist of Sandy and the platform a) The platform is initially tethered 1.00 m away from the right end of the pool. Ssndy initially at the left end of the platform, and walks with a constant velocity of 4.00 m/s towards the right. What is the kinetic energy of the system when Sandy reaches the right end of the platform? from the right end of the pool. Sandy again starts at the left end of the platform and wallks right with a constant velocity of 4.00 m/s relative to the platform. Now what is the kinetic energy of the system when Sandy reaches the right end of the platform? c) If there's a difference in your answers between Parts a and b, explain why that is so. Where did the energy go d) When Sandy reaches the right end of the platform in Part b, how far away is the platform from the right end of the pool? e) Sandy is now given a 10.0 kg lead weight to throw. On land, she can throw it 4.50 m. She again stands at the left end of the untethered platform, which is again initially 1.00 m away from the right end of the pool. She now throws the weight to the right. Exactly where does it land relative to the right end of the platform? Where does it land relative to the right end of the pool? f) 1.00 in 5.00 m 65.0 kg 110 kg Oil Figure 3 - Sandy the astronaut trains by walking on a platform floating frictionlessly on a pool of oil.

Explanation / Answer

3. m = 110 kg

l = 5 m

m' = 65 kg

a. as the platform is tethered, it does not move

so KE of the system = 0.5m'*4^2 = 520 J

b. let speed of platform be v while sandy walks on it

from conservation of moemntum

110*v = 65*4

u = 4 m/s

v = 2.3636 m/s

hecne KE of the system = 0.5mv^2 + 0.5m'u^2 = 827.272 J

c. the extra energy went in the motion of the platform which wasnt moving in case a

d. let distance of the platform form the right enfd of the pool when sandy reaches right end of the platform be x

then

(65*6 + 110*3.5)/(65 + 110) = (65*x + 110*(x + 2.5))/(65 + 110)

hence

x = 2.857 m

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