a 12.0 kg box is released from the top of smooth incline that is 5.00m long and makes an angle 40 degrees with the horizontal.(a)What will be the acceleration of the box?(b)how long will it take to reach the bottom?

Draw out a triangle diagram of the action. Mass X acceleration.

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To solve this problem, we can break it down into two parts: finding the acceleration of the box and determining how long it will take to reach the bottom of the incline.

(a) Finding the acceleration of the box:
We can start by determining the component of the gravitational force that acts along the incline. This component will cause the box to accelerate down the incline.

Step 1: Resolve the forces:
Since the incline is smooth, there is no frictional force acting on the box. The only force acting on the box is its weight (mg) which can be broken down into two components: one perpendicular to the incline and one parallel to it.

The component of the weight parallel to the incline can be calculated using the formula:

Force parallel to incline = m * g * sin(angle)

Here,
m = mass of the box = 12.0 kg
g = acceleration due to gravity = 9.8 m/s^2
angle = angle of incline = 40 degrees

Force parallel to incline = 12.0 kg * 9.8 m/s^2 * sin(40 degrees) ≈ 75.34 N

Step 2: Calculate the acceleration:
The acceleration is equal to the net force acting on the box divided by its mass. Since no other forces are acting on the box apart from the parallel component of the weight, the net force is equal to that component.

Acceleration = Force parallel to incline / mass = 75.34 N / 12.0 kg ≈ 6.28 m/s^2

Therefore, the acceleration of the box is approximately 6.28 m/s^2.

(b) Determining the time it takes to reach the bottom:
To find the time it takes for the box to reach the bottom of the incline, we can use the equation of motion that relates displacement, time, initial velocity, and acceleration:

s = ut + (1/2)at^2

Here,
s = displacement = 5.00 m (length of the incline)
u = initial velocity = 0 m/s (box starts from rest)
a = acceleration = 6.28 m/s^2 (calculated in part a)
t = time

Since the box starts from rest, the initial velocity is 0 m/s, and the equation simplifies to:

s = (1/2)at^2

Substituting the values:

5.00 m = (1/2) * 6.28 m/s^2 * t^2

Simplifying the equation:

t^2 = (2 * 5.00 m) / 6.28 m/s^2

t^2 ≈ 1.59 s

Taking the square root:

t ≈ √1.59 s ≈ 1.26 s

Therefore, it will take approximately 1.26 seconds for the box to reach the bottom of the incline.