The coefficient of kinetic friction between a floor and a box made of wood is µk = 0.13. How far does the box slide if it has an initial speed of v0 = 7.0 m/s?

s = 19.23 m

Explanation:

Since, the frictional force is given as:

f = μN = μW

f = μmg   equation (1)

but, from Newton's 2nd Law:

f = ma   equation (2)

comparing equation (1) and equation (2):

ma = μmg

a = μg

here,

a = acceleration of puck

μ = coefficient of static friction = 0.13

g = -9.8 m/s² (negative sign for slowing down of puck)

Therefore,

a = (0.13)(-9.8 m/s²)

a = -1.274 m/s²

Now, using 3rd equation of motion:

2as = Vf² - Vi²

where,

s = distance covered = ?

Vf = Final Velocity = 0 m/s (since, puck finally stops)

Vi = Initial Velocity = 7 m/s

Therefore,

2(- 1.274 m/s²)s = (0 m/s)² - (7 m/s)²

s = (- 49 m²/s²)/(- 2.548 m/s²)

s = 19.23 m

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards