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3 years ago

8

Starting from rest a 5.00 kg steel sphere rolls down a frictionless ramp with a height of 4.00 m. what is the spberes speed when

it reaches the bottom of the ramp
________m/s

A. 8.85

B. 78.4

C. 39.2

D. There is no energy lost from the system

1 answer:

kondor19780726 [428]3 years ago

3 0

Answer:

A. 8.85

Explanation:

U = mgh

5*4*9.8 = 0.5*5*V^2

V = √4*9.8/0.5 = √78.4 = 8.85 m/se

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$K_{A}+U_{Ag}=K_{B}+U_{Bs}+U_{Bg}+W_{Bf}$

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Now, for the second answer we need to find the acceleration of the elevator when it reaches point B, for which we need to build a free body diagram (also included in the attached picture) and to a summation of forces:

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$a=\frac{ky_{B}+f-W}{m}$

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$a=\frac{(9300N/m)(1m)+(17000N)-(2000kg)(9.8m/s^{2})}{2000kg}$

which yields:

$3.35m/s^{2}$

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