Question

A box of mass m1 rests on a smooth, horizontal floor next to a box of mass m2. Suppose the force of 20.0 N pushes on two boxes of unknown mass. We know, however, that the acceleration of the boxes is 1.20 m/s2 and the contact force has a magnitude of 4.45 N. Find the mass of box 1 and box 2.

Answer 1

Answer:

Explanation:

Given

acceleration of system a =1.2 m/s^2

Normal Force N=4.45 N

Force exerted F=20 N

Thus

$F=(m_1+m_2)a$

$\frac{20}{1.2}=m_1+m_2$

$16.67=m_1+m_2$-------1

Normal reaction $N=m_2a$

$4.45=m_2\times 1.2$

$m_2=3.70 kg$

therefore $m_1=16.67-3.70$

$m_1=12.96 kg$

Answer 2

Answer:

Mass of box 1 is 12.95 kg

Mass of box 2 is 3.7083 kg  

Explanation:

We have given the pushing force F = 20 N

Force on the box 2 that is contact force $F_{m2}=F_{c}=4.45N$

So net force that is force on box 1 $F_{m1}=F-F_c=40-4.45=15.55N$

It is given that acceleration for both the box is same as $a=1.20m/sec^2$

From newton's law we know that F = ma

So mass of box 1 $m_1=\frac{F_{m1}}{a}=\frac{15.55}{1.2}=12.9583kg$

Mass of box 2 $m_2=\frac{F_{m2}}{a}=\frac{4.45}{1.2}=3.7083kg$