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KatRina [158]
3 years ago
8

Two piston hydraulic problems

Physics
1 answer:
Solnce55 [7]3 years ago
8 0

Answer:

i forgot

Explanation:111

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D) All of the above

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Can someone help me ASAP
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A cord is used to vertically lower an initially stationary block of mass M = 3.6 kg at a constant downward acceleration of g/7.
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Answer:

(a) W_c=127.008 J

(b) W_g=148.176 J

(c) K.E. = 21.168 J

(d) v=3.4293m.s^{-1}

Explanation:

Given:

  • mass of a block, M = 3.6 kg
  • initial velocity of the block, u=0 m.s^{-1}
  • constant downward acceleration, a_d= \frac{g}{7}

\Rightarrow That a constant upward acceleration of \frac{6g}{7} is applied in the presence of gravity.

∴a=- \frac{6g}{7}

  • height through which the block falls, d = 4.2 m

(a)

Force by the cord on the block,

F_c= M\times a

F_c=3.6\times (-6)\times\frac{9.8}{7}

F_c=-30.24 N

∴Work by the cord on the block,

W_c= F_c\times d

W_c= -30.24\times 4.2

We take -ve sign because the direction of force and the displacement are opposite to each other.

W_c=-127.008 J

(b)

Force on the block due to gravity:

F_g= M.g

∵the gravity is naturally a constant and we cannot change it

F_g=3.6\times 9.8

F_g=35.28 N

∴Work by the gravity on the block,

W_g=F_g\times d

W_g=35.28\times 4.2

W_g=148.176 J

(c)

Kinetic energy of the block will be equal to the net work done i.e. sum of the two works.

mathematically:

K.E.= W_g+W_c

K.E.=148.176-127.008

K.E. = 21.168 J

(d)

From the equation of motion:

v^2=u^2+2a_d\times d

putting the respective values:

v=\sqrt{0^2+2\times \frac{9.8}{7}\times 4.2 }

v=3.4293m.s^{-1} is the speed when the block has fallen 4.2 meters.

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