# physics and forces?

Two blocks, m1 = 4.4 kg and m2 = 2 kg, are in contact with each other and are on a horizontal surface for which the coefficient of kinetic friction is 0.1. A 27 N horizontal force acts on block 1.What is the the force exerted by m2 on m1.?

I have tried every possible answer and none of them work, please help.

### 5 Answers

- WoodsmanLv 74 weeks agoBest answer
friction force

f = µ(m1+m2)g = 0.1 * 6.4kg * 9.8m/s² = 6.272 N

and so the net motive force

F = (27 - 6.272)N = 20.728 N

results in a system acceleration

a = F / (m1 + m2) = 20.728N / 6.4kg = 3.24 m/s²

In order to accelerate m2 at this rate, the applied force must be

F = m1*a + µ*m1*g = 2kg * (3.24 + 0.1*9.8)m/s² = 8.4 N

If the direction of motion is considered "positive," then the force that m2 exerts on m1 is equal and opposite, so you might need a negative sign.

- 4 weeks ago
Assuming that the 27 N is pushing m1 toward m2 then for the system as a whole: 27 - (.1)(6.4)(9.8) = 6.4 a then a = 3.23875 m/s*s. With m1 pushing on m2: P - (.1)(2)(9.8) = 2 a = 2(3.23875) Giving P = 8.4375 N.

- Anonymous4 weeks ago
Post your work.................

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- Born YesterdayLv 74 weeks ago
Well, unbalanced force results in acceleration.

The frictional force is (4.4+2)×.1×9.8 = 6.272N

So, the combined system should accelerate at

(27-6.272)÷6.6 = 2.181 m/s^2

2.181×2 + 2×.1×9.8 = 6.322 Newtons