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There is friction between the block and the
table.
The suspended 2 kg mass on the left is
moving up, the 4 kg mass slides to the right
on the table, and the suspended mass 4 kg on
the right is moving down.
The acceleration of gravity is 9.8 m/s
2
.What is the magnitude of the acceleration
of the system?
1. 1.67253
2. 1.9845
3. 1.75311
4. 2.47673
5. 2.1315
6. 2.058
7. 2.548
8. 1.33933
9. 1.5288
10. 2.03127
Answer in units of m/s
2
.

There is friction between the block and the table The suspended 2 kg mass on the left is moving up the 4 kg mass slides to the right on the table and the suspen class=

Respuesta :

phuyalrashik

Answer:

5 no

Explanation:

actually the 4kg lying on table has no influence

it slides towards 4kg weight hung

as it has excess 2kg force

force=miu × m ×g

snehashish65

The magnitude of acceleration of the entire system is  [tex]2.1315 \;\rm m/s^{2}[/tex]. Hence, option (5) is correct.

Given data:

The masses  suspended from the table are, m = 2 kg and m' = 4 kg.

And the mass kept on the table is, m'' = 4 kg.

The frictional force will oppose the motion of mass 4 kg, kept on the table. Then the required value of frictional force is,

[tex]f = \mu \times m'' \times g\\\\f = 0.11 \times 4 \times 9.8\\\\f = 4.312 \;\rm N[/tex]

Since, the mass kept on the right side is more. Then at equilibrium,

-f + m''g + (m' +m)a = m'g

Here,

a is the magnitude of acceleration of the system.

Solving as,

-4.312+ 4(9.8) + (4 -2)a = 4(9.8)

2a = 4.312

[tex]a \approx 2.1315 \;\rm m/s^{2}[/tex]

Thus, we can conclude that the magnitude of acceleration of the entire system is  [tex]2.1315 \;\rm m/s^{2}[/tex]. Hence, option (5) is correct.

Learn more about the linear acceleration here:

https://brainly.com/question/14214161

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