Answer: (a) The work done by this force on the particle is 42.71 J.
(b) The change in the potential energy of the system is -42.71 J.
(c) The kinetic energy the particle is 62.96 J.
Explanation:
(a) For the given situation, expression for work done is as follows.
W = [tex]\int_{0.9}^{5.15}Fdx[/tex]
= [tex]\int_{0.9}^{5.15}(2x + 4)dx[/tex]
= [tex][2\frac{x^{2}}{2} + 4x]^{5.15}_{0.9}[/tex]
= [tex](x^{2} + 4x)^{5.15}_{0.9}[/tex]
= [tex][(5.15)^{2} + 4(5.15) - (0.9)^{2} - 4(0.9)][/tex]
= 26.52 + 20.6 - 0.81 - 3.6
= 42.71 J
Hence, the work done by this force on the particle is 42.71 J.
(b) Expression for a conservative force is as follows.
F = [tex]-\frac{dU}{dx}[/tex]
dU = -Fdx
[tex]\int_{0.9}^{5.15}dU[/tex] = [tex]\int_{0.9}^{5.15}Fdx[/tex]
[tex]\int_{0.9}^{5.15}dU[/tex] = -42.71 J
Therefore, the change in the potential energy of the system is -42.71 J.
(c) According to the work energy theorem,
W = [tex]\Delta K.E[/tex]
[tex]K.E_{0.9} - K.E_{5.15}[/tex] = W
[tex]K.E_{0.9} = W + K.E_{5.15}[/tex]
= [tex]42.71 + \frac{1}{2}mu^{2}[/tex]
where, u = velocity of the mass at x = 0.9 m
u = 3.0 m/s, m = 4.50 kg
As, [tex]K.E_{0.9} = W + K.E_{5.15}[/tex]
= [tex]42.71 + \frac{1}{2} \times 4.50 \times (3)^{2}[/tex]
= 62.96 J
Therefore, the kinetic energy the particle is 62.96 J.
