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Determine the velocity of the 60-lb block A if the two blocks are released from rest and the 40-lb block B moves 2 ft up the incline. The coefficient of kinetic friction between both blocks and the inclined planes is Ī¼k=0.10.

Respuesta :

vā‚ = 0.771 ft/s

vb = -1.54 ft/s

Explanation:

Block A:

F = ma

Nā‚ = - 60 cos60Ā° = 0

Nā‚ = 30 lb

Fā‚ = 0.1 X 30 = 3lb

Block B:

F = ma

Nb = - 40 cos30Ā° = 0

Nb = 34.64lb

Fb = 0.1 X 34.64 = 3.464lb

T1 + āˆ‘U = T2

(0 + 0) + 60 sin 60Ā° |Ī”sā‚| - 40 sin 30Ā° |Ī”sb| - 3|Ī”sā‚| - 3.464 |Ī”sb|

= 1/2 (60/32.2) vā‚Ā² + 1/2 (40/32.2) vbĀ²

and

2vā‚ = - vb

On solving, we get

vā‚ = 0.771 ft/s

vb = -1.54 ft/s

Ver imagen thamimspartan
lhmarianateixeira

In this exercise we have to use the knowledge of mechanics, in order to calculate the value of the velocity of each block, so:

The velocity of A is Ā 0.771 ft/s and velocity of B is -1.54 ft/s.

How calcule the force of wich block?

So calculating the force that each block exerts is:

  • Block A:

[tex]F = ma\\N_a = - 60 cos60\Ā° = 0\\N_a = 30 lb\\F_a = 0.1 X 30 = 3lb[/tex]

  • Block B:

[tex]F = ma\\N_b = - 40 cos30\Ā° = 0\\N_b = 34.64lb\\F_b = 0.1 * 34.64 = 3.464lb[/tex]

By summing the Ā tension, we have that:

[tex]T_1 + \sum U = T_2\\(0 + 0) + 60 sin 60\Ā° |\Delta s_a| - 40 sin 30\Ā° |\Delta s_b| - 3|\Delta s_a| - 3.464 |\Delta s_b|\\= 1/2 (60/32.2) v_a^2 + 1/2 (40/32.2) v_b^2\\2v_a = - v_b[/tex]

On solving, we get

[tex]v_a = 0.771 ft/s\\v_b = -1.54 ft/s[/tex]

See more about velocity at brainly.com/question/16379705