Question
Asked Dec 18, 2019
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A force of magnitude Fx acting
in the x - direction on a 2.00 - kg
particle varies in time as shown
in Figure P6.16. Find (a) the
impulse of the force, (b) the
final velocity of the particle if
it is initially at rest, and (c) the
final velocity of the particle if it is
initially moving along the x - axis
with a velocity of -2.00 m/s.

F,(N)
4
t (s)
1 2 3 4
Figure P6.16
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F,(N) 4 t (s) 1 2 3 4 Figure P6.16

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Expert Answer

Step 1

 (a). The impulse delivered to the ball is equal to the area under the curve.

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impulse = F - At = area = (AN)(25) (4N)(2s)+(4N)(1s)+- = 12N:s %3D

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Step 2

(b). using impulse momentum theorem, the final velocity of the parti...

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F. Δt Δp = m(v, -v.) 12kg - m/s = (2kg)(v, -0) '; = 6m/s

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