A box slides across a frictionless floor with an initial speed of 1.8m/s. It encounters a rough region where the coefficient of friction is 0.7

a. 4.29 m b. 2.25 m/s

Explanation:

Here is the complete question

A box slides across a frictionless floor with an initial speed v = 2.9 m/s. It encounters a rough region where the coefficient of friction is µk = 0.1. a) What is the shortest length of rough floor which will stop the box? b) If instead the strip is only 1.71 m long, with what speed does the box leave the strip?

Solution

a) What is the shortest length of rough floor which will stop the box?

Using work-kinetic energy theorem,

ΔK = W where ΔK = kinetic energy change of box and W = work done by friction = -fd where f = frictional force and d = length of rough floor

So, ΔK = -fd  where f = μmg, μ = coefficient of friction = , m = mass of box and g = acceleration due to gravity = 9.8 m/s²

1/2m(v₂² - v₁²) = μmgd

v₁ = 2.9 m/s and v₂ = 0 m/s (since the box stops)

So,

1/2m(v₂² - v₁²) = -μmgd

(v₂² - v₁²)/2 = -μgd

d = -(v₂² - v₁²)/2μg

substituting the values of the variables into the equation, we have

d = -((0 m/s)² - (2.9 m/s)²)/(2 × 0.1 × 9.8 m/s²)

d = -(0 m²/s² - 8.41 m²/s²)/1.96 m/s²

d = -(- 8.41 m²/s²)/1.96 m/s²

d = 8.41 m²/s²/1.96 m/s²

d = 4.29 m

b) If instead the strip is only 1.71 m long, with what speed does the box leave the strip?

Using work-kinetic energy theorem,

ΔK = W where ΔK = kinetic energy change of box and W = work done by friction = -fd where f = frictional force and d' = length of strip = 1.71 m

So, ΔK = -fd  where f = μmg, μ = coefficient of friction = , m = mass of box and g = acceleration due to gravity = 9.8 m/s²

1/2m(v₂² - v₁²) = μmgd'

v₁ = 2.9 m/s and v₂ = unknown

So,

1/2m(v₂² - v₁²) = -μmgd'

(v₂² - v₁²)/2 = -μgd'

v₂² - v₁² = -2μgd'

v₂² = v₁² - 2μgd'

v₂ = √(v₁² - 2μgd')

substituting the values of the variables into the equation, we have

v₂ = √((2.9 m/s)²) - (2 × 0.1 × 9.8 m/s² × 1.71 m)

v₂ = √(8.41 m²/s² - 3.3516 m²/s²)

v₂ = √(5.0584 m²/s²)

v₂ = 2.25 m/s