Block 1 (2.5kg) is on the surface of a frictionless inclined plane that makes an angle of 20 degrees with the horizontal. Assume g = 10 m/s2
magnitude of the acceleration of block 1 down the incline plane is 3.4 m/s2
How far does block 1 travel down the plane in 1.5 s after it is released?
You already know the acceleration down the plan. They told you what it is.
a = 3.4 m/s^2.
The distance travelled in time t is
S = (1/2) a t^2
Just insert t = 1.5 s and compute your answer.
To determine how far block 1 travels down the plane in 1.5 seconds after it is released, we need to use the equations of motion.
First, let's define the variables:
- m1 represents the mass of block 1, which is stated as 2.5 kg.
- θ represents the angle of the inclined plane, which is given as 20 degrees.
- g represents the acceleration due to gravity, which is given as 10 m/s^2.
- a represents the acceleration of block 1 down the inclined plane, which is given as 3.4 m/s^2.
- t represents the time, which is given as 1.5 seconds.
- d represents the distance traveled by block 1 down the inclined plane.
Now, we can proceed with the solution.
The acceleration along the inclined plane can be calculated using the equation:
a = g * sin(θ)
Plugging in the values:
3.4 = 10 * sin(20)
To solve for sin(20), rearrange the equation:
sin(20) = 3.4 / 10
Next, calculate the distance traveled using the equation of motion for distance (d):
d = 0.5 * a * t^2
Plugging in the values:
d = 0.5 * (3.4 / 10) * (1.5)^2
Simplifying the equation:
d = 0.5 * 0.34 * 2.25
Calculating the value:
d ≈ 0.3825 meters
Therefore, block 1 travels approximately 0.3825 meters down the inclined plane in 1.5 seconds after it is released.