A force is applied to a particle of mass 0.590 kg, initially at rest. The force

Question

A force is applied to a particle of mass 0.590 kg, initially at rest. The force (Fx) as a function of displacementxis described by one of the graphs shown here. Determine, for each of these graphs, the particle's change in kinetic energy and velocity.
(Round answers to three significant figures; include units.)
HINT 1: Use the Work-Kinetic Energy Theorem. Think about how the information in the graph is related to the Work done by the force.
HINT 2: Recall that?v=vf-vi. The object is originally at rest and you can find the final velocity from the final Kinetic Energy.

Explanation / Answer

Work done = Change in KE = 1/2*mv^2

v = (2*KE/m)

Work done = Area under F-x plot

a) Area under plot = 0.4*1 = 0.4 N-m

KE = 0.4 J

v = (2*KE/m) = (2*0.4/0.59) = 1.164 m/s

b) Area under pplot = 1/2*0.5*2 = 0.5 N-m

KE = 0.5 J

v = (2*KE/m) = (2*0.5/0.59) = 1.302 m/s

c) Area under plot = 1/2*0.2*0.5 + 1/2*(0.2+0.5)*(2-0.5) = 0.575 n-m

KE = 0.575 J

v = (2*KE/m) = (2*0.575/0.59) = 1.396 m/s

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