A 2.5kg block is on 53° inclined plane for which coefficient of kinetic friction is 0.25 and coefficient of static friction is 0.5. Calculate its acceleration given that:
(a)it is intially at rest
(b)it is moving up the slope
M*g = 2.5 * 9.8 = 24.5 N. = Wt. of block.
Fp = 24.5*sin53 = 19.6 N. = Force parallel to the incline.
Fn = 24.5*Cos53 = 14.7 N. = Normal force.
Fs = us*Fn = 0.5 * 14.7 = 7.37 N.
Fk = uk*Fn = 0.25 * 14.7 = 3.69 N.
a. a = 0.
b. Fp-Fk = M*a.
a = (Fp-Fk)/M =
To calculate the acceleration of the block on the inclined plane, we will 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 = ma).
(a) When the block is initially at rest, the force of static friction will be acting to prevent the block from sliding down the incline.
To determine the maximum force of static friction, we need to find the normal force (N) and the coefficient of static friction (μs).
The normal force can be calculated using the formula N = mg * cos(θ), where m is the mass of the block, g is the acceleration due to gravity (approximately 9.8 m/s^2), and θ is the angle of the inclined plane.
N = (2.5 kg) * (9.8 m/s^2) * cos(53°)
N ≈ 19.93 N
The maximum force of static friction (Fs,max) can be found using the formula Fs,max = μs * N, where μs is the coefficient of static friction.
Fs,max = 0.5 * 19.93 N
Fs,max ≈ 9.97 N
Since the block is initially at rest, the maximum force of static friction will be equal to the force required to keep the block from slipping. Therefore, the net force acting on the block is zero.
Fs,max = m * a
9.97 N = (2.5 kg) * a
Solving for acceleration (a), we find:
a = 9.97 N / (2.5 kg)
a ≈ 3.99 m/s^2
Therefore, when the block is initially at rest, the acceleration of the block on the inclined plane is approximately 3.99 m/s^2.
(b) When the block is moving up the slope, the force of kinetic friction will be acting in the opposite direction of motion.
The force of kinetic friction (Fk) can be found using the formula Fk = μk * N, where μk is the coefficient of kinetic friction.
Fk = 0.25 * 19.93 N
Fk ≈ 4.98 N
The net force acting on the block will be the difference between the force component parallel to the incline (mg * sin(θ)) and the force of kinetic friction.
Net force (Fnet) = mg * sin(θ) - Fk
Fnet = (2.5 kg) * (9.8 m/s^2) * sin(53°) - 4.98 N
Using Newton's second law (F = ma), we can solve for acceleration (a):
Fnet = m * a
[(2.5 kg) * (9.8 m/s^2) * sin(53°) - 4.98 N] = (2.5 kg) * a
Solving for acceleration (a), we find:
a = [(2.5 kg) * (9.8 m/s^2) * sin(53°) - 4.98 N] / (2.5 kg)
a ≈ 2.65 m/s^2
Therefore, when the block is moving up the slope, the acceleration of the block on the inclined plane is approximately 2.65 m/s^2.