Question

A 4.0 kg box is initially at rest on a table. The box is pushed for 3.0 m with a horizonta force of 25 N. The frictional force between the box and table is 12 N. speed of the box after the box is pushed 3.0 meters. 2. Determine the Initial Final g s dis

Answer 1

Initial velocity (u) = 0
Now from the FBD
F - f = ma
Where F is applied force = 25 N
f is friction force = 12 N
m is mass of box = 4 kg
a is its acceleration
25 -12 =4a
a = 3.25 m/s²
Distance covered by the box(s) = 3 m
Now from the kinematic equation we can write
v² = u² +2as
where v is final velocity
v² = 0 +2*3.25*3
v = 4.416 m/s
Hence the final speed of the box would be 4.416 m/s

10 f= 12 N F=25 N