Sandy is training to spend time on the International Space Station by learning t

Question

Sandy is training to spend time on the International Space Station by learning to walk on a platform floating in a pool of oil. The platform has a mass of 110. kg, is 5.00 m long, and floats on the oil without any friction. Sandy, with all her gear on, has a total mass of 65.0 kg. We consider S the system T to consist of Sandy and the platform. The platform is initially tethered 1.00 in away from the right end of the pool. Sandy is 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? The tether is now untied so that the platform floats freely on the oil. starting 1.00m away from the right end of the pool. Sandy again starts at the left and of the platform and walks 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? If there's a difference in your answers between Parts a and b. explain why that is so. W here did the energy go? 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? Sandy is now given a 10.0 kg load weight to throw. On land, she can throw it 4.50 m. If she again stands at the left end of the untethered platform and throws the weight to the right, exactly where does it land relative to the right cud of the platform? Where does it land relative to the right end of the pool? Sandy the astronaut trains by walking on a platform floating frictionlessly on a pool

Explanation / Answer

a) First case

Kinetic energy of sandy is = 1/2 .65. 42 = 520 J

Kinetic energy of platform is 0

Total K.E = 520 J

b) Now

m V1 + M V2 = 0

65x 4 = -100 x V2

V2 = 2.6 m/s in other direction

Now K.E is

Kinetic energy of sandy is = 1/2 .65. 42 = 520 J

Kinetic energy of platform is = 1/2. 100.(2.6)2 =338 J

Total K.E = 858 J

C) The difference is the equations is the tension in the rope for stopping the platform to move the distance.

d) he walks at 4 m/s, so he covers 5 metres in 1.25 seconds

Distance travelled by platform = 2.6 x 1.25 = 3.25 meters

Total distance is 3.25 + 1 = 4.35 meters

e) The recoil distance is is 10 x 4.5 / (65+110) = 0.257

So the object lands . 4.5 + 0.257 = 4.757 meteres on the ground from left end

f) for Time of flight assuming astronaught is 1.5 m, t = 0.54 seconds. vx = 4.5/0.54 = 8.3 m/s

so m1V1 + m2V2 =0

(65+110) V1 = 10x 8.3

V1 = 83/175 = 0.47 m/s

Distance travelled is = 0.47 x 0.54 =.256 meters

distance from right end of pool is 1.5 meters approx

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