In the figure, a 5.40 kg block is sent sliding up a plane inclined at θ = 37.0° while a horizontal force Upper F Overscript right-arrow EndScripts of magnitude 50.0 N acts on it. The coefficient of kinetic friction between block and plane is 0.300. What are the (a) magnitude and (b) direction (up or down the plane) of the block's acceleration? The block's initial speed is 4.20 m/s. (c) How far up the plane does the block go? (d) When it reaches its highest point, does it remain at rest or slide back down the plane?
(a) magnitude: 2.90 m/s^2
(b) direction: up the plane
(c) How far up the plane does the block go: 8.45 m
(d) When it reaches its highest point, does it remain at rest or slide back down the plane: remain at rest
To find the magnitude and direction of the block's acceleration, we can start by calculating the net force acting on the block.
(a) The net force can be found using the equation:
Net force = applied force - frictional force
= Upper F Overscript right-arrow EndScripts - (coefficient of friction) * (normal force)
The normal force can be determined by taking the component of the weight force perpendicular to the plane:
Normal force = mass * gravity * cos(θ)
Using the given values, we can calculate the normal force:
Normal force = 5.40 kg * 9.8 m/s^2 * cos(37.0°) ≈ 41.83 N
Now, we can calculate the frictional force:
Frictional force = (coefficient of friction) * (normal force)
= 0.300 * 41.83 N ≈ 12.54 N
Substitute the values into the equation for the net force:
Net force = 50.0 N - 12.54 N ≈ 37.46 N
The magnitude of the block's acceleration is equal to the net force divided by the mass:
Acceleration = Net force / mass
= 37.46 N / 5.40 kg ≈ 6.93 m/s²
(b) The direction of the block's acceleration is determined by the net force. Since the applied force is directed up the plane and the frictional force acts in the opposite direction, the net force would be downward (opposite to the direction of the applied force). Therefore, the block's acceleration is down the plane.
(c) To determine how far up the plane the block goes, we can use the kinematic equation:
V^2 = V0^2 + 2as
The final velocity (V) at the highest point is 0 m/s since the block comes to rest. The initial velocity (V0) is 4.20 m/s, and the acceleration (a) is -6.93 m/s² (negative as it is down the plane). We need to determine the displacement (s).
Rearranging the equation, we have:
s = (V^2 - V0^2) / (2a)
= (0^2 - (4.20 m/s)^2) / (2 * -6.93 m/s²)
≈ -0.85 m
The negative sign indicates the displacement is in the opposite direction of the initial motion. Therefore, the block goes 0.85 meters down the plane.
(d) When the block reaches its highest point, it comes to rest (V = 0 m/s).
To solve this problem, we will break it down into different parts and use Newton's second law of motion, along with the concept of friction.
(a) To find the magnitude of the block's acceleration, we'll consider the forces acting on the block along the incline. These forces include the force applied (F_applied), the force of gravity (mg), the normal force (N), and the force of friction (f_kinetic). The equation we'll use is:
F_net = F_applied - f_kinetic - mg sin(θ) = ma
where m is the mass of the block.
First, calculate the force of friction:
f_kinetic = μ_k * N
where μ_k is the coefficient of kinetic friction.
Next, calculate the normal force:
N = mg cos(θ)
Now substitute these values back into the equation:
F_applied - μ_k * m * g - m * g * sin(θ) = m * a
Plug in the given values:
F_applied = 50.0 N
m = 5.40 kg
θ = 37.0°
μ_k = 0.300
g = 9.8 m/s^2 (acceleration due to gravity)
We'll solve this equation for acceleration (a).
(b) The direction of the block's acceleration depends on the net force acting on it. If the net force is positive, the acceleration is up the incline, and if the net force is negative, the acceleration is down the incline. So, we need to check whether the net force is positive or negative.
If the net force is positive, the block will move up the incline. If the net force is negative, the block will move down the incline.
(c) To find how far up the plane the block goes, we'll use the concept of work done by the net force. The work done is equal to the change in kinetic energy.
Work_done = ΔKE = KE_final - KE_initial
where KE = (1/2)mv^2
We'll consider the initial and final kinetic energy of the block to find the work done by the net force, and then use the work-energy theorem to solve for the distance traveled up the plane.
(d) We'll determine whether the block remains at rest or slides back down the plane when it reaches its highest point by considering the forces acting on it at that point. If the force of gravity pulling the block back down the incline is greater than the force opposing it, the block will slide back down. If the force opposing gravity is greater, the block will remain at rest.
Now let's go through the problem step by step to find the answers: (a), (b), (c), and (d).