Question

Two block of masses 7 kg and 5 kg are placed in
contact with each other on a smooth surface. If a
force of 6 N is applied on the heavier mass, the
force on the lighter mass is
​

Answer 1

Answer:

2.5 N

Explanation:

Since, surface is smooth, therefore both the blocks move with same acceleration (say a).

Common acceleration.

a=F/M+m

where F=6

    M= 7

    m=5

a=6/7+5

   =1/2 ms ^-2

therefore force on lighter mass = 2.5N

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Answer 2

The force on the lighter mass is 2.5 N

Given,

mass of the heavier block, m1 = 7kg

mass of the lighter block, m2 = 5kg

Force on the heavier block, F1 = 6N

To Find,

Force applied on the lighter block, F2

Solution,

We know that Force, F1 = (m1 + m2)a

Therefore, the common acceleration:

⇒ a = F1 ÷ (m1 + m2)

Substituting the given values, we get

⇒ a = 6 ÷ (7 + 5)

⇒ a = 6 ÷ 12

⇒ a = 0.5 m per second square

Now, Force on lighter mass, F2 = m2 × a

⇒ F2 = 5 × 0.5

⇒ F2 = 2.5 N

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