The magnitude of the net force exerted in the x direction on a 2.35-kg particle varies in time as shown in the figure below.

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The magnitude of the net force exerted in the x direction on a 2.35-kg particle varies in time as shown in the figure below.

image
(a) Find the impulse of the force over the 5.00-s time interval.
Iarrowbold = N · s

(b) Find the final velocity the particle attains if it is originally at rest.
varrowboldf = m/s

(c) Find its final velocity if its original velocity is -3.30 ihatbold m/s.
varrowboldf = m/s

(d) Find the average force exerted on the particle for the time interval between 0 and 5.00 s.
Farrowboldavg = N
asked Oct 23, 2011 in Physics by kirby ~Expert~ (3,020 points)
    

1 Answer

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Part a would just be the area under the curve, because I=N*s, and the graph is of newtons and seconds.

and the other parts can be solved using impulse momentum theorem.

net Impulse=change in momentum

and momentum=mass*velocity, using this information you should be able to solve for the other parts.
answered Jan 9, 2012 by Roar3215 ~Expert~ (1,318 points)
edited Jan 9, 2012 by Roar3215

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