Question

A 3.0kg and 5.0kg box rest side by side on a smooth, level floor. Ahorizontal force of 32N is applied to the 3.0 kg box pushing itagainst the 5.0 kg box and as a result both boxes slide along thefloor. How large is the contact force between the two boxes?

a) 12N
b) 20N
c) 32N
d) 0N

Answer 1

Answer:

a) 12N

Explanation:

The boxes move together, so we can use Newton's second law to find its acceleration

$F=ma$

in this case

$m=m_{1}+m_{2}$

where

$m_{1}=3kg$

and

$m_{2}=5kg$

thus the acceleration, given a force of 32N, is:

$a=\frac{F}{m}=\frac{F}{m_{1}+m_{2}}=\frac{32N}{3kg+5kg} =\frac{32N}{8kg} =4m/s^2$

and now we use this acceleration to find the force between the boxes, using the mass of the first box:

$F=m_{1}a=(3kg)(4m/s^2)=12N$

this is the force between the two boxes (box one on box two, and for the box two on box one is the same magnitude but in the opposite direction).