The only force acting on a 4.0 kg body as the body moves along an x axis varies

Question

The only force acting on a 4.0 kg body as the body moves along an x axis varies as shown in the figure. The scale of the figure's vertical axis is set by Fs = 8.0 N. The velocity of the body at x = 0 is 5.0 m/s. (a) What is the kinetic energy of the body at x = 4.0 m? (b) At what value of x will the body have a kinetic energy of 25.0 J? (c) What is the maximum kinetic energy of the body between x = 0 and x = 5.0 m? image url: http://edugen.wileyplus.com/edugen/courses/crs7165/art/qb/qu/ch0/EAT_13438513709340_02117753568200753.gif

Explanation / Answer

mass m=4 kg

Fs=8 N

at x=0 m ==> v=5 m/sec

a)

at x=0,

K.Ei=1/m*v^2 =1/2*4*5^2=50 J

work done from x=5m to 4m is,

w1=Fs*(dX)=-8*(4-5)=8 J

by using work-energy theorem,

kinetic energy at x=4m is,

Kf=Ki-w1

=50-8

=42J

b)

initial K.Ei=50 J

final K.E=K.Ei+(-Fs*(x-2)

K.Ef=K.Ei+(-Fs*(x-2)

25=50+(-8*(x-2))

x=5.12 m

c)

here,

kinetic energy is maximum at x=1 m,

and

work done work done from x=5m to 2m is,

w1=Fs*(dX)=-8*(2-5)=24 J

work done work done from x=2m to 1m is,

w2=1/2*Fs*(2-1)=1/2*8*(2-1)= 4 J

by applying work- energy theorem,

K.Emax=Ki-w1-w2

K.Emax=50-24-4=22 J

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