A 6 kg box with initial speed 8 m/s slides across the floor and comes to a stop after 2.4 s. A) What is the coefficient of kinetic friction?B) How far does the box move? C) You put a 5 kg block in the box, so the total mass is now 11 kg, and you launch this heavier box with an initial speed of 7 m/s. How long does it take to stop?

A. Coefficient of kinetic friction, μ = 0.34

B. The box moves a distance of 9.64 m before coming to a stop

C. The heavier box will stop after 2.1 seconds

Explanation:

a. The coefficient of kinetic or sliding friction is given as: μ = F/R

where applied force, F = m × (∆v)/t

∆v = v - u

∆v = 0 - 8 m/s = -8 m/s; t = 2.4 s

R = normal reaction = m×g

where g = 9.8 m/s²

Substituting in the kinetic friction formula; μ = m∆v/t ÷ 1/m×g

μ = ∆v/g×t

μ = 8 / 9.8 × 2.4

μ = 0.34

b. Using the equation v² = u² + 2as to calculate the distance travelled by the box

where v = 0 m/s; u = 8.0 m/s; a = ? s = ?

From F = ma = μR

a = μR/m = (μ × m × g)/m

a = μg

a = 0.34 × 9.8

a = 3.32 m/s²

This is negative acceleration or deceleration

Substituting in the equation of motion

8² + 2 × -3.32 × s = 0

-6.64s = -64

s = 9.64 m

Therefore, the box moves a distance of 9.64 m before coming to a stop

c. The coefficient of friction is independent of mass.

Using the formula in (a): μ = ∆v/g×t

t = ∆v/μg

t = 7/0.34 × 9.8

t = 2.10 s

Therefore, the heavier box will stop after 2.1 seconds

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2