A student, crazed by final exams, uses a force of magnitude 86 N and angle θ = 61° to push a 5.2 kg block across the ceiling of his room, as in Figure 6-52. If the coefficient of kinetic friction between the block and the ceiling is 0.34, what is the magnitude of the block's acceleration?
Well, it seems like this student is using all their force to push that block across the ceiling. But I have to say, pushing a block on the ceiling? That's really thinking outside the box!
Now, let's tackle this problem step by step. We have the force applied by the student, which is 86 N at an angle of 61°. Impressive force there, student!
To find the acceleration, we need to consider the forces acting on the block. Firstly, we have the force applied by the student, which we can break down into its horizontal and vertical components. The horizontal component would be 86 N * cos(61°), and the vertical component would be 86 N * sin(61°).
Now, since the block is on the ceiling, we mostly need to worry about the friction force opposing the motion. The coefficient of kinetic friction is given as 0.34.
The friction force is calculated as the coefficient of friction multiplied by the normal force. But since we're dealing with a block on the ceiling, the normal force would be equal to the weight of the block, which is 5.2 kg * 9.8 m/s^2 (acceleration due to gravity).
Once we have the friction force, we can use Newton's second law (F = ma) to find the acceleration. The net force on the block would be the force applied by the student minus the friction force.
Now, plug in the values and do the calculations, and you should get the magnitude of the block's acceleration. Just be careful not to make any "ceiling" jokes while doing the calculations; we don't want to distract the student!
To find the magnitude of the block's acceleration, we need to consider the forces acting on the block:
1. The student applies a pushing force, which can be broken down into horizontal and vertical components.
- The horizontal component of the force (F_horizontal) is calculated as F * cos(theta), where F is the magnitude of the force (86 N) and theta is the angle (61°).
- The vertical component of the force (F_vertical) is calculated as F * sin(theta).
2. The frictional force (F_friction) opposes the motion of the block and is calculated as the product of the coefficient of kinetic friction (μ_k) and the normal force (F_normal).
- The normal force (F_normal) is equal to the weight of the block, which is calculated as mass (m) multiplied by the acceleration due to gravity (g ≈ 9.8 m/s^2).
- F_normal = m * g, where m is the mass of the block (5.2 kg) and g is the acceleration due to gravity.
We can start by calculating the horizontal and vertical components of the force:
F_horizontal = F * cos(theta) = 86 N * cos(61°)
F_vertical = F * sin(theta) = 86 N * sin(61°)
Next, we calculate the normal force:
F_normal = m * g = 5.2 kg * 9.8 m/s^2
Finally, we calculate the magnitude of the block's acceleration:
Net force in the horizontal direction = F_horizontal - F_friction
Block's acceleration = (F_horizontal - F_friction) / mass
Let's plug in the values and calculate the acceleration.
To find the magnitude of the block's acceleration, we can use the equation of motion that relates force, mass, and acceleration.
1. First, let's resolve the force applied by the student into its x and y components.
The horizontal component of the force is given by: Fx = F * cos(θ)
The vertical component of the force is given by: Fy = F * sin(θ)
Plugging in the given values, we have:
Fx = 86 N * cos(61°)
= 86 N * 0.47
≈ 40.42 N
2. Now, let's consider the forces acting on the block.
The force of kinetic friction (fk) opposes the horizontal force applied by the student and is given by:
fk = μ * N
where μ is the coefficient of kinetic friction and N is the normal force.
The normal force is equal to the weight of the block acting vertically downward, which is given by:
N = m * g
where m is the mass of the block and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Plugging in the given values, we have:
N = 5.2 kg * 9.8 m/s^2
≈ 50.96 N
Therefore, the force of kinetic friction is:
fk = 0.34 * 50.96 N
≈ 17.33 N
3. Now, let's calculate the net force acting on the block horizontally.
The net force is the difference between the horizontal force applied by the student and the force of kinetic friction:
Fnet = Fx - fk
Plugging in the values, we have:
Fnet = 40.42 N - 17.33 N
≈ 23.09 N
4. Finally, we can use Newton's second law of motion to find the magnitude of the block's acceleration.
Newton's second law states: F = m * a
Rearranging the formula, we have:
a = F / m
Plugging in the values, we have:
a = 23.09 N / 5.2 kg
≈ 4.44 m/s^2
Therefore, the magnitude of the block's acceleration is approximately 4.44 m/s^2.
Wb = mg = 5.2kg * 9.8N/kg = 51N. =
Weight of block.
Fb = [51N,0deg).
Fp = 51sin(0) = 0 = Force parallel to
ceiling.
Fv=51cos(0) = 51N=Force perpendicular to ceiling.
Ff = u*Fv = 0.34 * 51 = 17.34N. = Force
of kinetic friction.
Fn = Fap*cosA - Fp - Ff,
Fn = 86cos61 - 0 - 17.34 = 24.4N. = Net force.
Fn = ma,
a = Fn / m = 24.4 / 5.2 = 4.68.