6. a 15 kg box is given an initial push so that it slides across the floor and comes to a stop.
if the coefficient of friction is 0.30,
fn=(.8) = 147.15
a) find the friction force.
(30 (142.15
b) find the acceleration of the box.
c) how far will the box go if its initial speed is 3.0 m/s?

(a) Frictional force = 44.145 N

(b) Acceleration = -2.943 m/s²

(c) Distance covered = 1.529 m

Explanation:

Given:

Mass of the box is,

Coefficient of friction is,

Acceleration due to gravity is,

Normal force acting on the box is,

(a)

Frictional force is given as:

Therefore, the frictional force is 44.145 N.

(b)

The acceleration of the box is given using Newton's second law as:

Therefore, the acceleration of the box is -2.943 m/s².

(c)

Initial velocity is,

Final velocity is,

Acceleration of the box is,

The displacement of the box can be determined using equation of motion as:

Therefore, the displacement of the box is 1.529 m.

the chart show the masses and velocities of two colliding objects that stick together after a collision.

according to the law of conservation of momentum, the momentum of the object after the collision is  

answer

1,500

kg • m/s.