A block of mass m = 1.70kg is released from rest h = 0.485m from the surface of a table, at the top of a θ = 27.4° incline as shown in the figure below.
I got a few of the questions
a)The frictionless incline is fixed on a table of height H = 1.98m. Determine the acceleration of the block as it slides down the incline.
answer:4.51 m/s^2
b) What is the velocity of the block as it leaves the incline?
answer: 3.08 m/s
questions need help with
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?
Wb = M*g = 1.70 * 9.8 = 16.7 N.
Fp = 16.7*sin27.4 = 7.67 N. = Force parallel to the incline.
a. a = Fp/M = 7.67/1.70 = 4.51 m/s^2.
b. PE = Mg*h = 16.7*0.485 = 8.08 J.
KE = 0.5M*V^2 = 8.08.
0.85*V^2 = 8.08.
V^2 = 9.51.
V = 3.08 m/s.
c.
To solve questions c) and d), we need to determine the time it takes for the block to hit the floor. Here's how we can find the answers:
c) How far from the table will the block hit the floor?
To find this distance, we need to calculate the horizontal displacement of the block when it hits the floor.
The horizontal displacement, denoted by "d," can be calculated using the equation:
d = v_initial * t + 0.5 * a * t^2
where:
- v_initial is the initial velocity of the block on the incline.
- a is the acceleration of the block calculated in question a).
- t is the time it takes for the block to hit the floor.
Since the block is released from rest, its initial velocity (v_initial) on the incline is 0. Therefore, the equation simplifies to:
d = 0.5 * a * t^2
To find "t," we can use the kinematic equation for vertical motion:
s = u * t + 0.5 * a * t^2
where:
- s is the vertical displacement (height of the table H),
- u is the initial vertical velocity of the block,
- a is the acceleration due to gravity.
Since the block is released from rest, its initial vertical velocity (u) is also 0.
Therefore, the equation becomes:
H = 0.5 * (-9.8 m/s^2) * t^2
Now we can solve for "t":
H = 4.9 * t^2
t^2 = H / 4.9
Substituting the given values:
t^2 = 1.98m / 4.9
t = sqrt(0.404s) = 0.635s
Now, we can calculate the horizontal displacement:
d = 0.5 * (4.51m/s^2) * (0.635s)^2
d = 0.907m
Therefore, the block will hit the floor approximately 0.907m away from the table.
d) How much time has elapsed between when the block is released and when it hits the floor?
We have already calculated the time it takes for the block to hit the floor, which is 0.635s. So, the time elapsed is 0.635 seconds.
To solve questions c) and d), we need to use the equations of motion and the kinematic equations.
c) To determine how far from the table the block will hit the floor, we need to find the horizontal distance traveled by the block. This distance is the horizontal component of the block's displacement.
First, we can find the time it takes for the block to hit the floor. We can use one of the kinematic equations to solve for time:
h = 0.5 * g * t^2
Where h is the vertical distance (0.485m), g is the acceleration due to gravity (9.8m/s^2), and t is the time.
Rearranging the equation to solve for time:
t = sqrt(2h / g)
t = sqrt(2 * 0.485 / 9.8)
t ≈ 0.313 seconds
Now that we know the time, we can find the horizontal distance traveled by the block:
x = v₀x * t
Where x is the horizontal distance, v₀x is the initial horizontal velocity component, and t is the time.
To find v₀x, we can use the initial velocity of the block at the top of the incline, which is the final velocity of the block at the bottom of the incline.
v₀x = vfx = v * cos(θ)
Where v is the final velocity of the block at the bottom of the incline (from question b) and θ is the inclination angle (27.4°).
Substituting the values:
v₀x = 3.08 * cos(27.4°)
v₀x ≈ 2.734 m/s
Now we can find the horizontal distance:
x = 2.734 * 0.313
x ≈ 0.856 meters
Therefore, the block will hit the floor approximately 0.856 meters away from the table.
d) To find the time that has elapsed from when the block is released till it hits the floor, we can use the time we previously found (0.313 seconds).
Therefore, the time elapsed is approximately 0.313 seconds.