Question

Blocks with masses of 3.0 kg
k
g
, 4.0 kg
k
g
, and 5.0 kg
k
g
are lined up in a row on a frictionless table. All three are pushed forward by a 11 N
N
force applied to the 3.0 kg
k
g
block. How much force does the 4.0 kg
k
g
block exert on the 5.0 kg
k
g
block?

Answer 1

The force exerted by the 4.0 kg block on the 5.0 kg block is 4.6 N

Explanation:

First of all, we need to find the acceleration of the system of three blocks together.

We can do it by applying Newton's second law of motion:

$F=Ma$

where:

F = 11 N is force applied to the system

M is the total mass of the system

a is its acceleration

The total mass of the 3 blocks is:

M = 3.0 + 4.0 + 5.0 = 12.0 kg

Therefore, the acceleration is:

$a=\frac{F}{M}=\frac{11}{12}=0.92 m/s^2$

Now we know that each of the 3 blocks has the same acceleration, $0.92 m/s^2$. We can therefore now apply Newton's second law again using only the block of 5.0 kg:

$F'=m_5 a$

where:

F' is the force exerted by the 4.0 kg block on the 5.0 kg block

$m_5 = 5.0 kg$ is the mass of the block

$a=0.92 m/s^2$ is its acceleration

And substituting,

$F'=(5.0)(0.92)=4.6 N$

Learn more about Newton's second law:

brainly.com/question/3820012

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