A block of mass m = 5.4 kg is pulled up a θ = 24° incline as in the figure with a force of magnitude F = 35 N.
(a) Find the acceleration of the block if the incline is frictionless.
m/s2
(b) Find the acceleration of the block if the coefficient of kinetic friction between the block and incline is 0.11.
m/s2
Oh, math questions! Strap on your thinking cap, because we're about to get calculating!
(a) If the incline is frictionless, then we only need to consider the force component that acts parallel to the incline. We can find that force by multiplying the force magnitude (35 N) by the sine of the angle (24°). So, F_parallel = 35 N * sin(24°) = 14.56 N.
Since there's no friction, this force will be responsible for accelerating the block. Using Newton's second law (F = ma), we can substitute the force with F_parallel and the mass with 5.4 kg to solve for acceleration:
14.56 N = 5.4 kg * a
Solving for a, we find:
a = 14.56 N / 5.4 kg ≈ 2.70 m/s²
So, the acceleration of the block is approximately 2.70 m/s².
(b) Now, let's consider the case where there is friction between the block and the incline. The force of kinetic friction can be found by multiplying the coefficient of kinetic friction (0.11) by the normal force. The normal force is equal to the weight of the block, which is m * g, where g is the acceleration due to gravity:
Normal force = 5.4 kg * 9.8 m/s² ≈ 52.92 N
Force of kinetic friction = 0.11 * 52.92 N ≈ 5.82 N
Next, let's break down the force parallel to the incline:
F_parallel = 35 N * sin(24°) = 14.56 N
Now, let's calculate the net force:
Net force = F_parallel - Force of kinetic friction = 14.56 N - 5.82 N = 8.74 N
Using Newton's second law once again, we can now solve for acceleration:
8.74 N = 5.4 kg * a
a = 8.74 N / 5.4 kg ≈ 1.62 m/s²
So, when factoring in the coefficient of kinetic friction, the acceleration of the block is approximately 1.62 m/s².
Phew! Math can be a bumpy ride, but we've made it through together.
To determine the acceleration of the block in both cases, we will use Newton's second law of motion, which states that the net force on an object is equal to the product of its mass and acceleration.
(a) If the incline is frictionless, the only force acting on the block is its weight, which can be resolved into two components: one parallel to the incline (mg*sin(θ)) and one perpendicular to the incline (mg*cos(θ)).
The net force in the parallel direction is given by: F_parallel = mg*sin(θ), where m is the mass of the block and g is the acceleration due to gravity (approximated as 9.8 m/s^2).
Using Newton's second law, we have: F_parallel = m*a, where a is the acceleration of the block.
Substituting the given values:
F_parallel = (5.4 kg)*(9.8 m/s^2)*sin(24°),
F_parallel ≈ 23.81 N.
Therefore, the acceleration of the block is given by:
a = F_parallel / m,
a ≈ 23.81 N / 5.4 kg,
a ≈ 4.41 m/s^2.
So, the acceleration of the block is approximately 4.41 m/s^2.
(b) If the coefficient of kinetic friction between the block and the incline is 0.11, there will be an additional force opposing the motion of the block, called the friction force.
The friction force can be calculated using the formula:
F_friction = coefficient of kinetic friction * normal force,
where the normal force is equal to mg*cos(θ).
The net force in the parallel direction is given by:
F_parallel = mg*sin(θ) - F_friction.
Using Newton's second law, we have:
F_parallel = m*a.
Substituting the given values and rearranging the equation, we get:
m*a = mg*sin(θ) - coefficient of kinetic friction * mg*cos(θ).
Simplifying further:
a = g*sin(θ) - coefficient of kinetic friction * g*cos(θ).
Substituting the values:
a ≈ (9.8 m/s^2)*sin(24°) - (0.11)*(9.8 m/s^2)*cos(24°),
a ≈ 3.45 m/s^2.
So, the acceleration of the block, when the coefficient of kinetic friction is 0.11, is approximately 3.45 m/s^2.
To solve both parts of the question, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the product of its mass and acceleration: F_net = m * a.
(a) When the incline is frictionless, the only force acting on the block is the component of the force F parallel to the incline, which we'll call F_parallel. The weight of the block (mg) is perpendicular to the incline and does not contribute to the acceleration.
To find F_parallel, we need to decompose the force F into its components. The component of F parallel to the incline can be found using the equation F_parallel = F * sin(θ), where θ is the angle of the incline.
F_parallel = 35 N * sin(24°) ≈ 14.63 N
Now we can use Newton's second law to find the acceleration:
F_net = m * a
F_parallel = m * a
Substituting the known values:
14.63 N = 5.4 kg * a
Dividing both sides by 5.4 kg, we find:
a ≈ 2.71 m/s^2
Therefore, the acceleration of the block on a frictionless incline is approximately 2.71 m/s^2.
(b) When there is kinetic friction between the block and the incline, we need to consider the friction force that opposes the motion. The magnitude of the friction force (F_friction) can be found using the equation F_friction = coefficient of kinetic friction * normal force, where the normal force (N) is equal to the weight of the block (mg) times the cosine of the incline angle.
N = mg * cos(θ)
N = 5.4 kg * 9.8 m/s^2 * cos(24°) ≈ 48.86 N
F_friction = 0.11 * 48.86 N ≈ 5.38 N
The net force acting on the block is now the component of F parallel to the incline minus the friction force:
F_net = F_parallel - F_friction
F_net = 14.63 N - 5.38 N ≈ 9.25 N
Using Newton's second law:
F_net = m * a
9.25 N = 5.4 kg * a
Dividing both sides by 5.4 kg, we find:
a ≈ 1.71 m/s^2
Therefore, the acceleration of the block on an incline with a coefficient of kinetic friction of 0.11 is approximately 1.71 m/s^2.
a) 35 - mgsin24 = ma
b) 35- mgsin24- Ff = ma
Fn = mgcos24
Ff = mu Fn