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

you cant move and you feel like someones pushing you down onto your bed it can feel like you want to get up but you just can't it can be extremely scary

acceleration = (change in speed) / (time for the change)

change in speed = (20-120)= -100 m/s.

1.5 min = 90 sec

acceleration = -100/90 = -1.11 m/s^2.

we don't care about the 150kg or the 2,500 meters. they don't make any difference.

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