A 3kg object moving along the x-axis has a velocity of 2.40m/s as it passes through the origin. It is acted on by a single force Fx that varies with x as shown...(adsbygoogle = window.adsbygoogle || []).push({});

a) find the work done by the force from x=0 to x = 2m

b) what is the kinetic energy of the object at x=2m?

c) what is the speed of the object at x=2m?

d) find the work done on the object from x=0 to x=4

e) what is the speed of the object at x =4m

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a) I know since the force is variable the work is equal to the area under the curve Fx-versus-x. I dont know wich value of F should I put in the Work integral equation S fx dx.

b) k = 1/2mv^2 =1/2(3kg)v^2

the force on the curve at x=2 is F = 1.9N can I put it the newton second law equation Ftotal = ma, 1.9N = 3kg x a ,

a = 1.9N/3kg = 0.63 m/s^2 and then since I got the acceleration I pluge it in v^2 = v + 2(a)(dleta x) = (2.40m/s)+2(0.63)(2m-0m)= 4.92

so v = 2.22m/s ?

d) before x=4m actually at x = 3m the curve go under the x-axis

I will integrate the part from x=0 to x=3 and then subtract from it the integration from x=3 to x = 4. but again I dont know what force should I use in the work integration fromula.

I thank you in advance of your help.

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# Homework Help: Work done by a variable force

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