A block of mass m = 2.00 kg is released from rest at h = 0.400 m from the surface of a table, at the top of a θ = 35.0° incline as shown below. The frictionless incline is fixed on a table of height H = 2.00 m.
(a) Determine the acceleration of the block as it slides down the incline.
m/s2
(b) What is the velocity of the block as it leaves the incline?
m/s
(c) How far from the table will the block hit the floor?
(D) How much time has elapsed between when the block is released and when it hits the floor?
To solve this problem, we can use principles of Newtonian mechanics. Let's break down the problem into smaller steps:
Step 1: Find the acceleration of the block as it slides down the incline.
We can start by finding the gravitational force acting on the block, which is given by the formula F = mg, where m is the mass of the block and g is the acceleration due to gravity (approximately 9.8 m/s^2).
So, F = (2.00 kg)(9.8 m/s^2) = 19.6 N.
Next, we need to calculate the component of the gravitational force parallel to the incline. This force is given by the formula F_parallel = mg*sin(θ), where θ is the angle of the incline (35.0°).
So, F_parallel = (2.00 kg)(9.8 m/s^2)*sin(35.0°) ≈ 10.67 N.
Now, we can find the net force acting on the block along the incline. Since the incline is frictionless, the net force is the same as the component of the gravitational force parallel to the incline.
Therefore, the net force (F_net) = 10.67 N.
Using Newton's second law of motion, F_net = ma, where a is the acceleration of the block.
Plugging in the values, we can find the acceleration:
10.67 N = (2.00 kg)a
a ≈ 5.34 m/s^2
So, the acceleration of the block as it slides down the incline is approximately 5.34 m/s^2.
Step 2: Find the velocity of the block as it leaves the incline.
To find the velocity, we can use the equation of motion: v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity (which is 0 in this case), a is the acceleration, and s is the displacement.
Since the block starts from rest, the initial velocity u = 0 m/s.
The displacement s is equal to the length of the incline, which can be found using trigonometry. The length of the incline L = h/sin(θ), where h is the vertical distance from the surface of the table to the top of the incline.
Plugging in the values, we get L = 0.400 m / sin(35.0°) ≈ 0.680 m.
Now, we can calculate the final velocity:
v^2 = 0^2 + 2*(5.34 m/s^2)*(0.680 m)
v ≈ 2.61 m/s
The velocity of the block as it leaves the incline is approximately 2.61 m/s.
Step 3: Find the distance from the table where the block hits the floor.
To find this distance, we need to calculate the horizontal distance traveled by the block after leaving the incline. Let's call this distance d.
Using the equation of motion, s = ut + (1/2)at^2, where s is the displacement, u is the initial velocity, a is the acceleration, and t is the time taken.
Since the block leaves the incline with a velocity v = 2.61 m/s, the initial velocity u for the horizontal motion is also 2.61 m/s.
The acceleration in the horizontal direction is 0 m/s^2 (assuming no external forces are acting in the horizontal direction).
Therefore, s = ut + (1/2)at^2 simplifies to d = ut, since (1/2)at^2 = 0.
Plugging in the values, we get d = (2.61 m/s) * t.
Step 4: Find the time taken for the block to hit the floor.
To find the time taken, we can use the equation of motion: v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time taken.
The final velocity (v) is the same as the velocity of the block as it leaves the incline, which is 2.61 m/s.
The initial velocity (u) is 0 m/s.
The acceleration (a) is the acceleration of the block as it slides down the incline, which is 5.34 m/s^2.
Plugging in the values, we get 2.61 m/s = 0 + (5.34 m/s^2) * t.
Solving for t, we get t ≈ 0.49 s.
So, the time taken for the block to hit the floor is approximately 0.49 seconds.
To summarize:
(a) The acceleration of the block as it slides down the incline is approximately 5.34 m/s^2.
(b) The velocity of the block as it leaves the incline is approximately 2.61 m/s.
(c) The horizontal distance from the table where the block hits the floor is given by d = (2.61 m/s) * t, where t is the time taken.
(d) The time elapsed between when the block is released and when it hits the floor is approximately 0.49 seconds.