To solve this problem, we'll need to break it down into different components. Let's start with the forces acting on the block:
The weight of the block acts vertically downward, given by: W = m * g where m is the mass of the block (2 kg) and g is the acceleration due to gravity (approximately 9.8 m/s^2).
The normal force (N) acts perpendicular to the incline. It cancels out the vertical component of the weight: N = m * g * cos(theta) where theta is the angle of the incline (30 degrees).
The force of friction (F_friction) opposes the motion of the block and acts parallel to the incline. Its magnitude is given by: F_friction = mu * N where mu is the coefficient of friction (0.3).
The force component parallel to the incline (F_parallel) helps to move the block up the incline and is given by: F_parallel = m * g * sin(theta)
Now, let's solve the problem:
a) How long does the block move up the plane?
To determine how long the block moves up the plane, we need to find the net force acting on the block and then use Newton's second law of motion (F = m * a) to calculate the acceleration.
Net force (F_net) = F_parallel - F_friction
Substituting the values: F_net = m * g * sin(theta) - mu * N = m * g * sin(theta) - mu * m * g * cos(theta)
Now, we can calculate the acceleration (a) using Newton's second law: F_net = m * a
Substituting the values: m * g * sin(theta) - mu * m * g * cos(theta) = m * a
Simplifying: g * sin(theta) - mu * g * cos(theta) = a
Substituting the known values: 9.8 m/s^2 * sin(30 degrees) - 0.3 * 9.8 m/s^2 * cos(30 degrees) = a
a ≈ 4.9 m/s^2
We have the acceleration (a). To find the time (t) taken to move up the plane, we can use the following kinematic equation:
v = u + a * t
Since the initial velocity (u) is given as 22 m/s and the final velocity (v) will be 0 m/s (as the block comes to a stop), we can solve for t:
0 = 22 m/s + 4.9 m/s^2 * t
Rearranging the equation: 4.9 m/s^2 * t = -22 m/s
t = -22 m/s / 4.9 m/s^2
t ≈ -4.49 seconds (We ignore the negative sign in this case)
Therefore, the block moves up the plane for approximately 4.49 seconds.
b) How far does the block move up the plane?
To find the distance (d) traveled by the block up the plane, we can use the following kinematic equation:
d = u * t + (1/2) * a * t^2
Substituting the known values: d = 22 m/s * 4.49 s + (1/2) * 4.9 m/s^2 * (4.49 s)^2
Simplifying: d ≈ 49.49 m
Therefore, the block moves up the plane for approximately 49.49 meters.