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

Question

A object of mass 3.00 kg is subject to a force Fx that varies with position as in the figure below. (a) Find the work done by the force on the object as it moves from x = 0 to x = 5.00 m J (b) Find the work done by the force on the object as it moves from x = 5.00 m to x = 8.00 m J (c) Find the work done by the force on the object as it moves from x = 10.0 m to x = 15.0 m. J (d) If the object has a speed of 0.500 m/s at x = 0, find its speed at x = 5.00 m and its speed at x = 15.0 m. speed at x = 5.00 m m/s speed at x = 15.0 m m/s

Explanation / Answer

Considering that this is your question image

http://www.webassign.net/sercp9/5-p-060.gif

Work = Force*distance = The area underneath the curve.

(a.)
Work done = (.5)(3)(1.75) = 2.625 J

(b.)
Work done = (3)(3) = 9 J

(c.)
Work done = (3)(5)(1/2) = 7.5 J

(d.)
From x = 0 to x = 5 m, the object had a constant acceleration of 3/5 m/s^2(The slope of the line). Therefore the speed of the object at 5 m would be:
vf = ?(0.5^2 + 2(3/5)(5)) = 2.5 m/s

From x = 10 to x = 15 m, the object had a constant acceleration of -3/5 m/s^2. Therefore, the speed of the object at 15 m would be:
vf = ?(2.5^2 + 2(-3/5)(5)) = 0.5 m/s

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