The following figure gives the force (Networks) versus position (meters) curve f

Question

The following figure gives the force (Networks) versus position (meters) curve for a force acting on a 1.5 - kg particle moving horizontally in one dimension. Compute the work done by the force F(x) as the particle answer from z1 = 2 meter to z2 = 18 meter Assuming that there is also indicate friction (with coefficient mu = 0.1) beside the force F(x) acting on this particle as it move along the x axis, determine the speed of the particle when it is located at zs = 18 meter, given that the speed of the particle where it is located at x1 = 2 meter is v1 = 5m/R.

Explanation / Answer

a)area from x=2 to x=16 gives us the work done by force f
=8+4-4-8+18 joules=18 joules

b)work done by friction=-.1*1.5*9.8*16j=-23.52J

total work done=-5.52
work done = change in kiinetic energy
-5.52=.5*1.5(v2 -52 )

v=4.2 m/s

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