In a pickup game of dorm shuffleboard, students crazed by final exams use a broom to propel a calculus book along the dorm hallway. If the 4.7 kg book is pushed from rest through a distance of 0.85 m by the horizontal 42 N force from the broom and then has a speed of 1.0 m/s, what is the coefficient of kinetic friction between the book and floor

μ_k = 0.851

Explanation:

We are given;

Mass of book; m_book = 4.7 kg

Horizontal force; F_horiz = 42 N

Distance; d = 0.85 m

Speed; v = 1 m/s

First of all let's find the acceleration using Newton's equation of motion;

v² = u² + 2ad

u is initial velocity and it's 0 m/s in this case.

Thus;

1² = (2 × 0.85)a

1 = 1.7a

a = 1/1.7

a = 0.5882 m/s²

Now, resolving forces along the vertical direction, we have;

W - N = 0

Thus,W = N

Where W is weight = mg and N is normal force

Thus; N = mg = 4.7 × 9.81 = 46.107 N

Now, resolving forces along the horizontal direction, we have;

F_horiz - ((μ_k)N) = ma

Where μ_k is coefficient of kinetic friction.

Thus;

42 - 46.107(μ_k) = 4.7 × 0.5882

42 - 46.107(μ_k) = 2.76454

μ_k = (42 - 2.76454)/46.107

μ_k = 0.851

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