You place a book of mass 5.00 kg against a vertical wall. You apply a constant force F to the book, where F = 92.0 N and the force is at angle of 60.0 above the horizontal. The coefficient of kinetic friction between the book and the wall is 0.300.
If the book is initially at rest, what is its speed after it has traveled 0.400 m up the wall.
Well, well, well, looks like the book is going on a little adventure up the wall! Let's calculate its speed, shall we?
First, let's find the net force acting on the book. We have the force you applied, F = 92.0 N, which makes an angle of 60.0 degrees with the horizontal. To find the vertical component of this force, we can use some trigonometry: F_vertical = F * sin(60.0) = 92.0 N * sin(60.0) = 79.4 N.
The force of friction is given by the coefficient of kinetic friction, μ_k, times the normal force, which is the weight of the book. The weight of the book is m * g, where m is the mass of the book (5.00 kg) and g is the acceleration due to gravity (9.8 m/s^2). So, the force of friction is F_friction = μ_k * m * g = 0.300 * 5.00 kg * 9.8 m/s^2 = 14.7 N.
The net force acting on the book is the force you applied minus the force of friction, so F_net = F_vertical - F_friction = 79.4 N - 14.7 N = 64.7 N.
Now, let's use Newton's second law, which states that the net force equals the mass times the acceleration (F_net = m * a). We know the net force (64.7 N) and the mass of the book (5.00 kg), so we can solve for the acceleration: a = F_net / m = 64.7 N / 5.00 kg = 12.94 m/s^2.
Finally, we can use the kinematic equation v^2 = u^2 + 2as to find the final speed (v) of the book. The initial speed (u) is 0 m/s since the book was initially at rest, and the acceleration (a) is 12.94 m/s^2. The distance traveled (s) is 0.400 m. Plugging in the values, we get v^2 = 0^2 + 2 * 12.94 m/s^2 * 0.400 m = 10.35 m^2/s^2.
Taking the square root of both sides, we find that the final speed of the book is v = sqrt(10.35 m^2/s^2) ≈ 3.22 m/s.
So, after traveling 0.400 m up the wall, the book will be zipping along at approximately 3.22 m/s. Have fun, little book!
To solve this problem, we'll use Newton's second law, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration:
F_net = m * a
We'll consider the forces acting on the book along the vertical direction:
1. The weight of the book (mg): This force is given by the product of the mass and the acceleration due to gravity (9.8 m/s^2).
2. The normal force (N): This force acts perpendicular to the vertical wall and cancels out the vertical component of the applied force.
3. The force of kinetic friction (fk): This force opposes the motion of the book along the wall and is given by the product of the coefficient of kinetic friction (μk) and the normal force (N).
Let's break down the forces involved:
1. The vertical component of the applied force (F_v): This force acts opposite to the weight and can be calculated using trigonometry:
F_v = F * sin(θ) = 92.0 N * sin(60°)
2. The net force in the vertical direction (F_net): This is the difference between the vertical component of the applied force and the weight:
F_net = F_v - mg
3. The net force in the horizontal direction (F_horizontal): This force is responsible for the horizontal acceleration:
F_horizontal = F * cos(θ) = 92.0 N * cos(60°)
Now that we've determined the forces involved, let's calculate the net force and acceleration:
1. Net force in the vertical direction:
F_net = F_v - mg
= (92.0 N * sin(60°)) - (5.00 kg * 9.8 m/s^2)
2. Net force in the horizontal direction:
F_horizontal = F * cos(60°)
Next, we'll calculate the acceleration along the vertical direction:
a = F_net / m
Now, using the kinematic equation relating displacement (Δy), initial velocity (v₀), acceleration (a), and time (t):
Δy = v₀ * t + (1/2) * a * t²
Since the book starts from rest, the initial velocity (v₀) is zero. We're given the displacement (Δy) as 0.400 m and want to find the final velocity (v).
Therefore, we can rearrange the equation to solve for v:
v² = 2 * a * Δy
v = sqrt(2 * a * Δy)
Finally, substitute the calculated values for a and Δy into the equation to find the final velocity (v) of the book after traveling 0.400 m up the wall.