An object slides down a frictionless 33 degree incline whose vertical height is 54.0 cm. How fast is it going in meters/second when it reaches the bottom?

KE gained = PE lost

(1/2)mv^2 = mgh
v = sqrt(2gh)= sqrt(2*9.8*0.54)m/s

First of all convert 54cm to m

1m=100cm, therefore 54cm=0.54m=h. then using equation of a motion v^2=u^+2gh,since u =0 g=10 the equation bcoms v^2=2gh then substitute and solve for v ans=32.9

To find the speed of the object when it reaches the bottom of the incline, we can use the principles of conservation of energy.

Step 1: Calculate the gravitational potential energy (PE) of the object at the top of the incline using the formula: PE = mgh, where m is the mass, g is the acceleration due to gravity (9.8 m/s^2), and h is the vertical height of the incline (54.0 cm = 0.54 m).

PE = mgh = m * 9.8 m/s^2 * 0.54 m

Step 2: Determine the angle of the incline in radians (θ) using the formula: θ = π/180 * degree, where π is approximately equal to 3.14159 and the degree is 33.

θ = π/180 * 33 = 0.575958653 radians

Step 3: Calculate the kinetic energy (KE) of the object at the bottom of the incline using the formula: KE = 1/2 * m * v^2, where v is the velocity of the object at the bottom of the incline.

Step 4: Apply the conservation of energy principle, which states that the total mechanical energy (PE + KE) of the object is conserved. Thus, PE = KE.

Equating the potential energy (PE) at the top of the incline to the kinetic energy (KE) at the bottom of the incline, we can solve for v.

m * 9.8 m/s^2 * 0.54 m = 1/2 * m * v^2

Simplifying the equation, we can cancel out the mass (m) on both sides:

9.8 m/s^2 * 0.54 m = 1/2 * v^2

Step 5: Solve for v by isolating it on one side of the equation.

4.90 m^2/s^2 * 0.54 m = v^2

2.646 m^2/s^2 = v^2

Taking the square root of both sides, we get:

v = √(2.646 m^2/s^2)

v ≈ 1.626 m/s

Therefore, the object will be going approximately 1.626 m/s when it reaches the bottom of the incline.

To find the speed of the object when it reaches the bottom of the incline, we can use the principles of physics and trigonometry. Let's break down the problem into steps:

Step 1: Calculate the vertical distance (height) covered by the object.
Given: vertical height = 54.0 cm

We need to convert cm to meters since we want the final answer in meters/second.
1 meter = 100 centimeters

height = 54.0 cm * (1 meter / 100 cm) = 0.54 meters

Step 2: Use the trigonometric function to find the horizontal distance covered by the object.
Given: Angle of the incline = 33 degrees
We can use the sine function to calculate the horizontal distance covered.

sine(angle) = height / hypotenuse
hypotenuse = height / sine(angle)

hypotenuse = 0.54 meters / sin(33 degrees) ≈ 1.028 meters

Step 3: Use the equation of motion to find the final velocity of the object when it reaches the bottom.
Since the object slides down a frictionless incline, we can assume there is no loss of energy.
We can use the equation:

v^2 = u^2 + 2as

where v is the final velocity, u is the initial velocity (which is zero in this case since the object starts from rest),
a is the acceleration, and s is the displacement.

The acceleration can be calculated using the formula:

a = g * sin(angle)

where g is the acceleration due to gravity.

Plugging in the values:

a = 9.8 m/s^2 * sin(33 degrees) ≈ 5.28 m/s^2
s = hypotenuse = 1.028 meters

v^2 = 0 + 2 * 5.28 m/s^2 * 1.028 meters
v^2 ≈ 10.87 m^2/s^2

Taking the square root of both sides:

v ≈ √(10.87 m^2/s^2)
v ≈ 3.30 m/s

Therefore, the object will be going approximately 3.30 meters/second when it reaches the bottom of the incline.