A box slides down a frictionless 35 degree incline. Determine its acceleration. If the incline is 10.0 m long and the box starts from
rest, how much time does it take to get to the bottom?
To determine the acceleration of the box sliding down the incline, we can use the following equation:
acceleration = g * sin(theta)
Where:
g = acceleration due to gravity = 9.8 m/s^2
theta = angle of the incline = 35 degrees
Let's calculate the acceleration:
acceleration = 9.8 * sin(35)
acceleration ≈ 5.64 m/s^2
Now, to find the time it takes for the box to reach the bottom of the incline, we can use the following equation:
distance = initial velocity * time + (1/2) * acceleration * time^2
Since the box starts from rest (initial velocity = 0), the equation simplifies to:
distance = (1/2) * acceleration * time^2
The distance is given as 10.0 m, so we can plug in the values and solve for time:
10.0 = (1/2) * 5.64 * time^2
Rearranging the equation:
time^2 = (10.0 * 2) / 5.64
time^2 ≈ 3.55
Taking the square root of both sides:
time ≈ √3.55
time ≈ 1.88 s
Therefore, it takes approximately 1.88 seconds for the box to reach the bottom of the incline.
To determine the box's acceleration, we can use Newton's second law of motion, which states that the force acting on an object is equal to its mass multiplied by its acceleration (F = m*a). In this case, the only force acting on the box is the component of its weight parallel to the inclined plane.
The weight of an object can be calculated by multiplying its mass (m) by the acceleration due to gravity (g). The acceleration due to gravity is approximately 9.8 m/s^2.
The component of the weight parallel to the inclined plane can be found by multiplying the weight (m*g) by the sine of the angle of the incline (sinθ). In this case, the angle is 35 degrees.
So, the formula for the force parallel to the incline is:
F = m * g * sinθ
Since this force is causing the acceleration, we can set it equal to m*a:
m * g * sinθ = m * a
We can cancel out the mass (m) on both sides of the equation:
g * sinθ = a
Now we can substitute the values:
a = (9.8 m/s^2) * sin(35 degrees)
a ≈ 5.68 m/s^2 (rounded to two decimal places)
Therefore, the acceleration of the box down the incline is approximately 5.68 m/s^2.
To find the time it takes for the box to reach the bottom of the incline, we can use the kinematic equation:
v = u + a * t
where v is the final velocity, u is the initial velocity (which is 0 because the box starts from rest), a is the acceleration, and t is the time.
Since the final velocity (v) at the bottom of the incline is unknown, and the initial velocity (u) is 0, we can simplify the equation to:
v = a * t
Rearranging the equation to solve for time (t):
t = v / a
The final velocity (v) can be found using the equation:
v^2 = u^2 + 2 * a * s
Where u is the initial velocity (0 m/s), a is the acceleration (5.68 m/s^2), and s is the distance traveled (10.0 m).
Plugging in the values:
v^2 = 0 + 2 * (5.68 m/s^2) * (10.0 m)
v^2 = 113.6 m^2/s^2
Taking the square root of both sides to solve for v:
v ≈ 10.67 m/s (rounded to two decimal places)
Now we can substitute the values into the equation for time (t):
t = v / a
t = (10.67 m/s) / (5.68 m/s^2)
t ≈ 1.88 seconds (rounded to two decimal places)
Therefore, it takes approximately 1.88 seconds for the box to reach the bottom of the incline.
component of gravity force down slope = m g sin 35
so
a = g sin 35
10 = (1/2) a t^2
20 /(g sin 35) = t^2