Question

three blocks of masses 2kg, 3kg, and 5 kg are placed in contact as shown in the diagram. a horizontal force of 40 n is applied on the 2kg block amd 10.n.is.applied on the 5 kg block . find n2 and n1. plz solve and post... ​

Answer 1

Answer:

Refer to the attachments for the Free Body Diagram.

Considering First Block having a mass of 2kg:

Applying the Newton's second law, we get:

⇒ ∑ F = Mass × Acceleration

Let us assume that the block system is moving with an acceleration of 'a' m/s². Hence applying the second law, we get:

⇒ 40 - N2 = Mass of Block × Acceleration

⇒ 40 - N2 = 2a

⇒ 40 - 2a = N2   ...(1)

Considering Block 2 having mass of 3 kg:

⇒ N2 - N1 = 3a

Substituting value of N2 from (1) we get:

⇒ 40 - 2a - N1 = 3a

⇒ 40 - 2a - 3a = N1

⇒ 40 - 5a = N1   ...(2)

Considering Block 3 having mass of 5 kg:

⇒ N1 - 10 = 5a

Substituting N1 from (2) we get:

⇒ 40 - 5a - 10 = 5a

⇒ 30 = 5a + 5a

⇒ 30 = 10 a

⇒ a = 30/10 = 3 m/s²

Hence the acceleration of the whole system is 3 m/s².

The value of N1 and N2 are:

• N1 = 40 - 5a
• N1 = 40 - 5(3) = 40 - 15 = 25 N

• N2 = 40 - 2a
• N2 = 40 - 2(3) = 40 - 6 = 34 N

Hence the values are N1 = 25 N and N2 = 34 N.

Hence Option (2) is the correct answer.

Answer 2

Answer:

Correct option is (b)

Explanation:

Given:

Mass,

$m_{1} =2kg\\m_{2} =3kg\\m_{3} =5kg$

Force applied on blocks, F=40N

According to Newton's Law of Motion:

Force = mass * acceleration

Let the Acceleration be 'a'.

This acceleration will be the same on all three blocks.

$a_{1} =a_{2} =a_{3} =a$

Force applied on Block 1

$40-N_{2} =m_{1} a\\40-N_{2}=2a ......(1)$

Similarly, Force applied on Block 2

$N_{2} -N_{1}=3a......(2)$

Similarly, Force applied on Block 3

$N_{1} -10=5a......(3)$

By simplifying equations (1), (2), and (3), we get:

$N_{1}=25N\\ N_{2}=34N$

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