Anonymous
Anonymous asked in Science & MathematicsPhysics · 1 month ago

# 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.?

Relevance
• 1 month ago

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.

• 1 month 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.

• Anonymous
1 month ago

• oubaas
Lv 7
1 month ago

F(2→1) = -27*2/(6.4) = -8.44 N