Answer:
a. 4.29 m b. 2.25 m/s
Step-by-step 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