A 2.0 x 103 kg car is pulled 345 m up a hill that make s an angle of 15° with the horizontal. What is the potential energy of the car at the top of the hill? If the car rolls down the hill, what will its speed be if we neglect friction? plz help? confused

How high is the hill? Answer: 345*sin15

PE= mgh

it is 654 muiles ahead

To calculate the potential energy of the car at the top of the hill, we can use the following formula:

Potential Energy (PE) = mass (m) x gravitational acceleration (g) x height (h)

First, let's find the height of the hill using trigonometry. The vertical component of the hill is given by:

Vertical component = distance x sin(angle)
= 345 m x sin(15°)

Next, we can substitute the given values into the formula:

PE = 2.0 x 10^3 kg x 9.8 m/s^2 x (345 m x sin(15°))

Calculate the value within brackets:

PE = 2.0 x 10^3 kg x 9.8 m/s^2 x (345 m x 0.259)

Simplify further:

PE = 2.0 x 10^3 kg x 9.8 m/s^2 x 89.355

Now, solve for PE:

PE = 17616660 J (rounded to the nearest whole number)

Therefore, the potential energy of the car at the top of the hill is approximately 17,616,660 Joules.

Moving on to the second part of the question, neglecting friction, the potential energy at the top of the hill will be converted into kinetic energy as the car rolls down. To calculate the speed of the car, we can use the formula for kinetic energy:

Kinetic Energy (KE) = (1/2) x mass (m) x velocity squared (v^2)

Since we are neglecting friction, we can assume that all of the potential energy is converted into kinetic energy:

PE = KE

Therefore:

PE = (1/2) x m x v^2

Rearranging the equation to solve for velocity:

v^2 = (2 x PE) / m

Substitute the values:

v^2 = (2 x 17,616,660 J) / 2.0 x 10^3 kg

Simplify:

v^2 = 35,233.32 m^2/s^2

Take the square root of both sides:

v = √(35,233.32 m^2/s^2)

Solve for v:

v ≈ 187.72 m/s (rounded to the nearest whole number)

Therefore, the speed of the car as it rolls down the hill neglecting friction is approximately 188 m/s.