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.
(b) Find the work done by the force on the object as it moves from x = 5.00 m to x = 11.0 m.
(c) Find the work done by the force on the object as it moves from x = 11.0 m to x = 16.0 m.
(d) If the object has a speed of 0.600 m/s at x = 0, find its speed at x = 5.00 m and its speed at x = 16.0 m.
speed at x = 5.00 m _______ m/s
speed at x = 16.0 m _______ m/s
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Work done = Area under F-x curve .
a) Work from x = 0 to 5 = Area under the triangle = 0.5 * 5 * 3 = 7.5 J.
b) Work from x = 5 to 11 = Area of rectangle formed by drawing vertical lines from x = 5 and x= 11.
Base of rectangle = 11-5 = 6
Height = 3.
Work = Area = 3*6 = 18 J.
c) Work from x= 11 to 16 = area under triangle = 0.5 * 5 * 3 = 7.5 J.
d) Work done = change in Kinetic energy.
At x=0, K.E. = 0.5 * m * v^2 = 0.5 * 3 * 0.36 = 0.54 J .
At x= 5, K.E. = 0.5 * m * v^2.
Therefore 0.5 * 3 * v^2 - 0.54 = 7.5
0.5*3*v^2 = 8.04
At x= 5, v = 2.31 m/s
At x = 11m, K .E. be K.E.1.
Change in K.E. = further work done that is 18J.
K.E.1 - 8.04 (K.E. at x=5) = 18
K.E.1 = 26.04
At x= 16m, K.E be K.E.2
K.E.2- KE1 = 7.5
KE2 = KE1+ 7.5 = 26.04+7.5 = 33.54
0.5*3*v^2 = 33.54
At x = 16 m , v = 4.72 m/s.
A object of mass 3.00 kg is subject to a force Fx that varies with position as in the figure below.
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