4) A single conservative force acts on a 5.00 kg particle within a system due to

Question

4) A single conservative force acts on a 5.00 kg particle within a system due to its interaction with the rest of the system. The equation Fx = 2x + 4 describes the force, where Fx is in newtons and x is in meters. As the particle moves along the x axis from x = 1.00 m to x = 5.00 m, calculate
a) the work done by this force on the particle,
b) the change in the potential energy of the system, and
c) the kinetic energy the particle has at x = 5.00 m if its speed is 3.00 m/s at x = 1.00 m.

5) In an experiment, one of the forces exerted on a proton is Fx = ?x2 in the x-direction, where ? = 12 N/m2.
a) How much work does Fx do when the proton moves from x = 0.10 m to x = 0.20 m?
b) From x = 0.20 m to x = 0.70 m?
c) From x = 0.70 m to x = 0.10 m?
d) Is the force Fx conservative? If Fx is conservative, what is the potential energy function Ux for it? Let Ux = 0 when x = 0.

6) A block of mass 1.6 kg is attached to a horizontal spring that has a spring constant k of 1 x 103 N/m. The spring is compressed 2.0 cm and is then released from rest on a frictionless surface.
a) Calculate the speed and elastic potential energy of the block as it passes through the equilibrium position x = 0. (Elastic potential energy means potential energy associated with a stretched or compressed spring.)
b) Calculate the speed and elastic potential energy of the block when it is in the position x = 1.1 cm.

Explanation / Answer

1. a) Fdx = (2x+4)dx
integrated w/r/t x yields
W = x^2 + 4x
evaluated from 1.0 to 5.0 m:
W = 5^2 + 4*5- 1^2 - 4*1 = 40 J

b) -40 J (force is conservative)
c) dU = -dK (change in potential = change in kinetic)
Ki = 0.5*m*v^2 = 0.5*5.0kg*(3m/s)^2 = 22.5J
Kf = Ki + dK = 22.5 J + 40J = 62.5 J

2. question is not clear or unable to read your given data

3.Initial Energy
E = 1/2 k d

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