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
At the bottom, 1/2 m v^2= PE at top
To find the potential energy of the car at the top of the hill, we can use the equation:
Potential Energy (PE) = mgh
where:
m = mass of the car = 2.0 x 10^3 kg
g = acceleration due to gravity = 9.8 m/s^2
h = height of the hill
Since the height is not directly given, we can calculate it using the given information. The height (h) can be found using the trigonometric relationship between the angle and the height of the hill. In this case, the opposite side of the triangle is the height (h), and the hypotenuse is the distance (d) the car is pulled up the hill.
Using the trigonometric equation:
sin(angle) = opposite/hypotenuse
sin(15°) = h/345 m
Now, we can solve for h:
h = sin(15°) * 345 m
Calculate the value of h using a calculator:
h ≈ 89.18 m
Now, we can substitute the values back into the equation to find the potential energy:
PE = (2.0 x 10^3 kg) * (9.8 m/s^2) * (89.18 m)
Calculating this expression:
PE ≈ 1.75 x 10^6 J
So, the potential energy of the car at the top of the hill is approximately 1.75 x 10^6 Joules.
To determine the speed of the car when rolling down the hill, we can use the conservation of energy principle. At the top of the hill, all the potential energy is converted into kinetic energy at the bottom of the hill.
The formula for kinetic energy (KE) is:
KE = 1/2 * m * v^2
where:
m = mass of the car = 2.0 x 10^3 kg
v = velocity of the car
Since potential energy is converted entirely into kinetic energy:
PE = KE
1/2 * m * v^2 = m * g * h
Canceling out the 'm' on both sides of the equation:
1/2 * v^2 = g * h
Now, solving for v:
v^2 = 2 * g * h
Substituting the known values:
v^2 = 2 * (9.8 m/s^2) * (89.18 m)
Calculating this expression:
v^2 ≈ 3,475.12 m^2/s^2
Finally, find the square root of both sides to get the velocity:
v ≈ √(3,475.12 m^2/s^2)
Calculating on a calculator:
v ≈ 58.99 m/s
Therefore, neglecting friction, the speed of the car when rolling down the hill is approximately 58.99 m/s.