Question

A 1 kg object is moving with a speed of 1.80 m/s in the x direction when it passes the origin, as shown in the figure below. It is acted on by a single force F_x that varies with x.

x, IT -2 -3

(a) What is the work done by the force from x = 0 to x = 2 m? (Note: You will not be able to calculate the answer directly, but there is a way to estimate it!)
J

(b) What is the kinetic energy of the object at x = 2 m?
J

(c) What is the speed of the object at x = 2 m?
m/s

(d) What is the work done on the object from x = 0 to x = 4 m?
J

(e) What is the speed of the object at x = 4 m?


I got a and b correct, but dont get on c, d, and e

Answer 1

(a)

work done equals to area under curve

W = Area= between the x =0 to x = 2 m we have 22 square

each square area has energy 0.125 J

W =22 ( 0.125 J ) = 2.75 J

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(b)

the kinetic energy of the object at x = 2 m is

K = Ki + W = (1/2) (1) ( 1.8)^2 +2.75 J = 4.37 J

(c)

the speed of the object at x = 2 m is

K = 1/2 * m v^2

v = root 2K/m = root 2 (4.37)/1.0 kg = 2.95 m/s

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(d)

the work done on the object from x = 0 to x = 4 m is

W = Area= between the x =0 to x = 2 m we have 28 square

each square area has energy 0.125 J

W = 28 ( 0.125 J) = 3.5 J

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he kinetic energy of the object at x = 4 m is

K = Ki + W = (1/2) (1) ( 1.8)^2 +3.5= 5.12 J

v = root 2K/m = root 2 ( 5.12 J/ 1.0kg = 2.26 m/s