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Vedmedyk [2.9K]
3 years ago
12

A 65-kg ice skater stands facing a wall with his arms bent and then pushes away from the wall by straightening his arms. At the

instant at which his fingers lose contact with the wall, his center of mass has moved 0.45m , and at this instant he is traveling at 3.0m/s .
A. What are the average force exerted by the wall on him?.
B. What are the work done by the wall on him?.
C. What are the change in the kinetic energy of his center of mass?
Physics
1 answer:
Hitman42 [59]3 years ago
8 0

Answer:

A) F=650N

B) W=0J

C) \Delta K=292.5J

Explanation:

A) Considering the equation v_f^2=v_i^2+2ad, we can calculate:

F=ma=m\frac{v_f^2}{2d}=(65kg)\frac{(3m/s)^2}{2(0.45m)}=650N

B) The work the wall does is 0J, because the fingers of the scater against it do not move. It's the muscles on his arm that do the work.

C)\Delta K=K_f-K_i=\frac{mv_f^2}{2}-\frac{mv_i^2}{2}=\frac{(65kg)(3m/s)^2}{2}-\frac{(65kg)(0m/s)^2}{2}=292.5J

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Read 2 more answers
Two subway stops are separated by 1210 m. If a subway train accelerates at 1.30 m/s2 from rest through the first half of the dis
solong [7]

Answer:

Part 1) Time of travel equals 61 seconds

Part 2) Maximum speed equals 39.66 m/s.

Explanation:

The final speed of the train when it completes half of it's journey is given by third equation of kinematics as

v^{2}=u^2+2as

where

'v' is the final speed

'u' is initial speed

'a' is acceleration of the body

's' is the distance covered

Applying the given values we get

v^2=0+2\times 1.30\times \frac{1210}{2}\\\\v^{2}=1573\\\\\therefore v=39.66m/s

Now the time taken to attain the above velocity can be calculated by the first equation of kinematics as

v=u+at\\\\v=0+1.30\times t\\\\\therefore t=\frac{39.66}{1.30}=30.51seconds

Since the deceleration is same as acceleration hence the time to stop in the same distance shall be equal to the time taken to accelerate the first half of distance

Thus total time of journey equalsT=2\times 30.51\approx61seconds

Part b)

the maximum speed is reached at the point when the train ends it's acceleration thus the maximum speed reached by the train equals 39.66m/s

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