A object of mass 3.00 kg is subject to a force F_x that varies with position as

Question

A object of mass 3.00 kg is subject to a force F_x that varies with position as in the figure below. Find the work done by the force on the object as it moves from x = 0 to x = 5.00 m. Find the work done by the force on the object as it moves from x = 5.00 m to x = 10.0 m. Find the work done by the force on the object as it moves from x = 10.0 m to x = 17.0 m. If the object has a speed of 0.400 m/s at x = 0, find its speed at x = 5.00 m and its speed at x = 17.0 m. speed at x = 5.00 m m/s speed at x = 17.0 m m/s

Explanation / Answer

given graph is F vs x
a)   work done is W= F*x
   so work done by force between x= 0 - 5 m is area under the curve that is
   area of the triangle
   = 1/2*5*2 (1/2*b*h)
    = 5 J
b) x from 5 - 10 m
   area of the rectangle with 4m , 5 m sides
   = 4*5 = 20 J
c)   feom x= 10 - 17 m
   area of the triangle
   = 1/2*7*4
   = 14 J
d)       at x= 5 m the work done = 5J = k.e
   5= 1/2 m*(v2^2-v1^2) work energy theorem
5= 1/2 *3(v2^2-0.16)
v2= 1.868 m/s
       at x= 17 m
39= 1/2 3*(v2^2-0.16)
v2= 5.08 m/s

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