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 = 3.00 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 = 11.0 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

Work = Force*distance = The area underneath the curve. (a.) Work done = (6/5)(2)(1/2) = 1.2 J (b.) Work done = (1)(3) = 3 J (c.) Work done = (4)(-3/5*(11) + 9)(1/2) = 4.8 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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