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4 years ago

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Two blocks are in contact on a frictionless, horizontal tabletop. An external force, F ,is applied to block 1, and the two block

s are moving with a constant acceleration of 2.70 m/s2. Use M1 = 3.03 kg and M2 = 6.40 kg. (a) What is the magnitude, F, of the applied force? (b) What is the contact force between the blocks? (c) What is the net force acting on block 1? {25.5 N, 17.3 N, 8.18 N to the right}

1 answer:

Zigmanuir [339]4 years ago

3 0

Answer:

Explanation:

The blocks are moving in the x direction

Given that,

m1=3.03kg

m2=6.40kg

After a force F is applied to the two masses they move with an acceleration of 2.7m/s²

Newton's Second Law:

ΣF = Σ(ma) = m1a1 + m2a2

Therefore

But in this case a1 =a2=2.7m/s², because they are connected by a string and they are accelerating at the same speed.

F=(m1+m2)a

F=(3.03+6.40)2.7

F=9.43 ×2.7

F=25.46N.

b. The contact force be blocks

F2= m2a

F2 is the force between the blocks

F2=6.4×2.7

F2= 17.28N

c. Fnet on block 1

It is the net force pulling the two blocks forward minus the force between the block 1 and 2

F1=F-F2

F=25.46-17.28

F=8.18N

The net force on block 1 is 8.18N

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