Question

The 1.23kg physics book shown is connected by a string to a 505.0g coffee cup. The book is given a push up the slope (call this direction positive) and released with a speed of 2.21m/s. The coefficients of friction are μs = 0.450 and μk = 0.221.


a) What is the acceleration of the book if the slope is inclined at 30.0°?

b) How far does the book slide before it has zero speed?

Answer 1

(a) The book decelerates as it moves upwards with magnitude of 6.77 m/s²

(b) The distance traveled by the book before stopping is 0.36 m.

The given parameters;

• mass of the book, m = 1.23 kg
• mass of coffee cup, m₁ = 505 g = 0.505 kg
• initial velocity, u = 2.21 m/s
• coefficient of kinetic friction, μk = 0.221

The vertical component of the force on the book-cup system;

$F_n = mg\ cos(\theta)$

The frictional force on the system;

$F_k = \mu_k F_n\\\\F_k = \mu_k .\ mg\ .cos(\theta)$

The horizontal component of the force on the system;

$-mg\ sin(\theta) - F_k = ma\\\\-mg\ sin(\theta) -\mu_kmg \ cos(\theta) = ma\\\\-g\ sin(\theta) -\mu_kg \ cos(\theta) = a\\\\-g[sin(\theta) +\mu_k \ cos(\theta)]= a$

The acceleration of the book along the slope is calculated as;

$a = -g[sin(\theta) + \mu_k\ cos(\theta)]\\\\a = -9.8[sin(30) + 0.221\times cos(30)]\\\\a = -9.8(0.691)\\\\a = -6.77 \ m/s^2$

Thus, the book decelerates as it moves upwards with magnitude of 6.77 m/s²

(b) The distance traveled by the book before it comes to stop is calculated as;

$v^2 = u^2 + 2as\\\\0 = (2.21)^2 + (2)(-6.77)(s)\\\\13.54 s = 4.8841 \\\\s = \frac{4.884}{13.54} \\\\s = 0.36 \ m$

Thus, the distance traveled by the book before stopping is 0.36 m.

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