The 3.42 kg physics book shown is connected by a string to a 518.0 g coffee cup. The book is given a push up the slope and released with a speed of 2.73 m/s. The coefficients of friction are μs = 0.490 and μk = 0.185. What is the acceleration of the book if the slope is inclined at 27.4°?
To find the acceleration of the book, we need to consider the forces acting on it.
Let's break down the forces into components parallel and perpendicular to the slope:
1. Determine the force due to gravity acting parallel to the slope:
F_parallel = m_book * g * sin(θ)
where m_book is the mass of the book, g is the acceleration due to gravity (9.8 m/s²), and θ is the angle of the slope (27.4°).
F_parallel = 3.42 kg * 9.8 m/s² * sin(27.4°)
2. Determine the force due to gravity acting perpendicular to the slope:
F_perpendicular = m_book * g * cos(θ)
F_perpendicular = 3.42 kg * 9.8 m/s² * cos(27.4°)
3. Determine the force of friction acting on the book. Since the book is at rest initially, we need to use the static friction coefficient:
F_friction = μs * F_perpendicular
where μs is the static friction coefficient.
F_friction = 0.490 * (3.42 kg * 9.8 m/s² * cos(27.4°))
4. Determine the net force acting on the book:
F_net = F_parallel - F_friction
5. Finally, use Newton's second law of motion (F = m * a) to find the acceleration:
F_net = m_book * a
a = F_net / m_book
Plug in the known values and solve for a:
a = (3.42 kg * 9.8 m/s² * sin(27.4°) - 0.490 * (3.42 kg * 9.8 m/s² * cos(27.4°))) / 3.42 kg
Simplify the equation:
a = (3.42 * 9.8 * sin(27.4°) - 0.490 * (3.42 * 9.8 * cos(27.4°)))
Calculate the value:
a ≈ 6.16 m/s² (rounded to two decimal places)
So, the acceleration of the book is approximately 6.16 m/s².