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

answer: at 24 february, 2014 the population gets twice of its initial population.

explanation:

since we have given that

the population for alpha city, t years after january 1, 2004 is given as

first we find out the initial population, i.e. t=0,

so, our quadratic equation becomes,

according to question, we have said that if the population has twice its initial population, then it becomes

so, time taken to reach this above mentioned population is given by

now, we know that time can't be negative so, we take t=9.149 years.

hence, at 24 february, 2014 the population gets twice of its initial population.