Question

Free-Body Diagrams 1 2 3 Net Force on Each Body Use your force diagrams to find Fnet for each body. Finet = N F2net N F3net N Mass X Acceleration of Each Body Use your measured values of masses and acceleration to find ma of each body. mi Xa х m2 X az m3 X 3 =

Force probe 1=0.47, force probe 2= 0.93, Gravitational force body 3=0.98

m1=0.66kg, m2=0.65kg, m3=0.1kg

a=0.7

Answer 1

Here,
Force probe 1 $\rightarrow F_{1}=0.47\ N$
Force probe 2
+F2=0.93 N

Gravitational force
F3 0.98 N

Mass of block 1
mi 0.66 kg

Mass of block 2
m2 = 0.65 kg

Mass of block 3 $\rightarrow m_{3}=0.1\ kg$
Acceleration $\rightarrow a=0.7\ m/s^{2}$
Net force in each block
net

Since, block is accelerating in positive x direction, net force $F_{1net}$ in block 1 in horizontal direction by applying Newton's laws of motion:

$\therefore F_{1net}=m_{1}a$

$\therefore F_{1net}=0.66 \times0.7$

${\color{Blue} \therefore F_{1net}=0.462\ N}$

Therefore, the net force on block 1 is 0.462 N.

Since, block 2 is accelerating in positive x direction, net force $F_{2net}$ in block 2 in horizontal direction by applying Newton's laws of motion:

$F_{2net}=-F_{1}+F_{2}$

$\therefore F_{2net}=-0.47+0.93$

*2net = mga 0.46 N

Therefore, the net force on block 2 is 0.46 N.

Since, block 3 is accelerating in negative y direction, net force $F_{3net}$ in block 3 in vertical direction by applying Newton's laws of motion:

$F_{3net}=F_{2}-F_{3}$

$\therefore F_{3net}=0.93-0.98$

${\color{Blue} \therefore F_{3net}=m_{3}a=-0.05\ N}$

Therefore, the net force on block 3 is -0.05 N.

Here, negative sign indicates the downward net force direction.