Two blocks are attached to either end of a light rope that passes over a light, frictionless pulley suspended from the ceiling. One block has mass 8.00 kg, and the other has mass 6.00 kg. The blocks are released from rest.
a.) For a 0.200 m downward displacement of the 8.00 kg block, what is the change in the gravitational potential energy associated with each block? Express your answers in joules separated by a comma.
Δ U 6, ΔU8 =
b.)If the tension in the rope is T, how much work is done on each block by the rope? Express your answers in terms of T separated by a comma.
W 6, W8 =
c.) Apply conservation of energy to the system that includes both blocks. During the 0.200 m downward displacement, what is the total work done on the system by the tension in the rope? What is the change in gravitational potential energy associated with the system? Use energy conservation to find the speed of the 8.00 kg block after it has descended 0.200 m. Express the energy in joules. Express the speed in meters per second. Enter your answers separated by commas
W, ΔUg, v2 = ________J, J, m/s

Respuesta :

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]

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