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# Answer to Question #5365 in Mechanics | Relativity for Damany Wallace

Question #5365
Physics help (Kinetic Energy) A bead with mass (1.8x10^-2)kg is moving along a wire in the positive direction of an x axis. Beginning at time=0, when the bead passes through x=0 with speed 12m/s, a constant force acts on the bead, the picture indicates the bead&#039;s position at these four times: t0=0 sec., t1=1.0 sec., t2=2.0 sec., t3=3.0 sec., The bead momentarily stops at t=3.0 sec. What&#039;s the kinetic energy of the bead as t=10 sec? Do I use the K=(.5)mv^2 formula or the vf^2=vi^2+2ad formula, THEN the K-formula?
1
2011-12-01T08:43:11-0500
1. You can find the first acceleration imparted to the bead constant force.
2. After 3 second the bead will move in another way with speed
3. After 10 second the speed of the bead is
4. Kinetic energy

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