Answer:
Part a)
[tex]\Delta U_6 = 11.77 J[/tex]
[tex]\Delta U_8 = -15.7 J[/tex]
Part b)
[tex]W_6 = T(0.200)[/tex]
[tex]W_8 = -T(0.200)[/tex]
Part c)
[tex]W_t = 0[/tex]
[tex]\Delta U = -3.93 J[/tex]
[tex]v = 0.75 m/s[/tex]
Explanation:
Part a)
change in potential energy is given as
[tex]\Delta U = mg(H_2 - H_1)[/tex]
now for 6 kg block we have
[tex]\Delta U_6 = (6)(9.81)(0.200)[/tex]
[tex]\Delta U_6 = 11.77 J[/tex]
for 8 kg block
[tex]\Delta U_8 = (8)(9.81)(-0.200)[/tex]
[tex]\Delta U_8 = -15.7 J[/tex]
Part b)
Work done by tension on 6 kg block is given as
[tex]W_6 = T(0.200)[/tex]
Work done by tension on 8 kg block is given as
[tex]W_8 = -T(0.200)[/tex]
Part c)
Total work done by the tension force is given as
[tex]W_t = 0[/tex]
total change in potential energy of the system is given as
[tex]\Delta U = -15.7 + 11.77[/tex]
[tex]\Delta U = -3.93 J[/tex]
Also by energy conservation we have
[tex]W = \Delta U + \Delta K[/tex]
[tex] 0 = -3.93 + \frac{1}{2}(8 + 6) v^2[/tex]
[tex]v = 0.75 m/s[/tex]