A 2.0-kg body moving along the x axis has a velocity vx = 5.0 m/s at x = 0. The

Question

A 2.0-kg body moving along the x axis has a velocity vx = 5.0 m/s at x = 0. The only force acting on the object is given by Fx = (-4.0x) N, where x is in m. For what value of x will this object first come (momentarily) to rest?
(I calculated 6.4 but apparently it's not the answer)

4.2 m
3.5 m
5.3 m
6.4 m
5.0 m

In a given displacement of a particle, its kinetic energy increases by 25 J while its potential energy decreases by 10 J. Determine the work of the nonconservative forces acting on the particle during this displacement.
(I chose -15 because it's the force by the NON-conservative force)
-15 J
+35 J
+15 J
-35 J
+55 J

Explanation / Answer

1. Energy in = Energy out KE = Work .5mv^2 = F x d .5(2)(5^2) = 4d d=6.25m For some reason, thats not one of the options...check to see if one of the variables you gave in the problem are incorrect. 2. Total energy in the system is constant So if we have +25 and -10 = 15 joules left, so work must perform -15 joules of work Hope this helps!

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