A car is sitting at the top of an inclined plane, which is 5.2 meters long and meets the horizontal at an angle of 12 degrees. The cart is then allowed to roll to the bottom of the incline. What will be the velocity of the cart at the bottom of the incline. (4.61m/s)

Please help me!!

I tried to solve it:
mgh=1/2mv^2
(resulted in incorrect answer?)

(1/2) m v^2 = m g h

so
v^2 = 2 g h

h = 5.2 sin 12
so
v^2 = 2 * 9.8 * 5.2 * sin 12

v^2 = 21.2
v = 4.60

To find the velocity of the cart at the bottom of the incline, you can use the principle of conservation of energy. The potential energy at the top of the incline is converted into kinetic energy at the bottom.

The potential energy (PE) of an object at a certain height (h) is given by the equation: PE = mgh, where m is the mass, g is the acceleration due to gravity (approximately 9.8 m/s²), and h is the height.

The kinetic energy (KE) of an object is given by the equation: KE = 1/2mv², where m is the mass and v is the velocity.

Since the potential energy is converted into kinetic energy, you can equate the two equations: PE = KE.

So, mgh = 1/2mv².

Now, let's calculate the potential energy at the top of the incline:
PE = mgh = m * 9.8 m/s² * 5.2 m * sin(12°).
Note: We use sin(12°) because the angle between the incline and horizontal is 12 degrees.

To simplify the equation, we can cancel the mass (m) on both sides:
gh = 1/2v².

Now, let's solve for the velocity (v):
v² = 2gh.
v = √(2gh).

Now, plug in the values:
v = √(2 * 9.8 m/s² * 5.2 m * sin(12°)).

Calculating this expression will yield the velocity of the cart at the bottom of the incline.