Question

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=(15)(-9.8) = 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?

Answer 1

Answer:

(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, $m=15\ kg$

Coefficient of friction is, $\mu =0.30$

Acceleration due to gravity is, $g=9.81\ m/s^2$

Normal force acting on the box is, $N=mg=(15)(9.81) = 147.15\ N$

(a)

Frictional force is given as:

$f=\mu N=0.30\times 147.15=44.145\ N$

Therefore, the frictional force is 44.145 N.

(b)

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

$a=-\frac{f}{m}=-\frac{44.145}{15}=-2.943\ m/s^2$

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

(c)

Initial velocity is, $u=3.0\ m/s$

Final velocity is, $v=0\ m/s$

Acceleration of the box is, $a=-2.943\ m/s^2$

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

$v^2=u^2+2ad\\0=3^2+2\times (-2.943)d\\0=9-5.886d\\5.886d=9\\d=\frac{9}{5.886}=1.529\ m$

Therefore, the displacement of the box is 1.529 m.