The cart starts from rest at the top of a hill with height, H1. It rolls down the hill and at the bottom of the hill it has a speed of 8 m/s. Next it is at the top of same hill but pushed so that it starts at the top of the hill with a speed of 6 m/s. How fast is it going at the bottom of the hill?

The change in kinetic energy will be the same in both cases.

(M/2)8^2 = M/2[V^2 - 6^2]

cancel the M's

V = sqrt[6^2 + 8^2] = 10 m/s

To find the speed of the cart at the bottom of the hill in the second scenario, we can use the principle of conservation of mechanical energy.

The mechanical energy of an object is the sum of its kinetic energy (KE) and potential energy (PE). In this case, as the cart is on a hill, the potential energy is given by the formula:

PE = m * g * h

where m is the mass of the cart, g is the acceleration due to gravity, and h is the height of the hill.

At the top of the hill, the cart has only potential energy, so we can write:

PE1 = m * g * H1 (Equation 1)

At the bottom of the hill, the cart has only kinetic energy, so we can write:

KE2 = 1/2 * m * v2^2 (Equation 2)

where v2 is the speed of the cart at the bottom of the hill in the second scenario.

Since the mechanical energy is conserved, we can equate the initial mechanical energy to the final mechanical energy:

PE1 = KE2

m * g * H1 = 1/2 * m * v2^2

Simplifying and canceling out the mass on both sides of the equation, we get:

g * H1 = 1/2 * v2^2

Now we can solve for v2:

v2^2 = 2 * g * H1

Taking the square root of both sides, we have:

v2 = √(2 * g * H1)

where g is the acceleration due to gravity (approximately 9.8 m/s²) and H1 is the height of the hill.

Thus, to find the speed of the cart at the bottom of the hill in the second scenario, you can plug in the values of g and H1 into the formula:

v2 = √(2 * 9.8 * H1)

To determine the speed of the cart at the bottom of the hill in both scenarios, we can apply the principle of conservation of energy. According to the law of conservation of energy, the total mechanical energy of an object remains constant as long as no external forces such as friction act on it.

Let's break down the problem into two parts:

Scenario 1:
In this scenario, the cart starts from rest at the top of the hill. The potential energy of the cart due to its height, H1, is converted into kinetic energy at the bottom of the hill. Using the conservation of energy, we can equate the potential energy at the top to the kinetic energy at the bottom.

Potential energy at the top = Kinetic energy at the bottom.
mgh1 = (1/2)mv^2

Where:
m is the mass of the cart
g is the acceleration due to gravity (approximately 9.8 m/s^2)
h1 is the height of the hill
v is the speed of the cart at the bottom of the hill

Rearranging the equation, we can solve for v:

v = sqrt(2gh1)

Substituting the values, we get:

v = sqrt(2 * 9.8 * H1)

Scenario 2:
In this scenario, the cart starts at the top of the hill with a speed of 6 m/s. This means it already has some kinetic energy in addition to the potential energy at the top. Again, applying the conservation of energy:

Potential energy at the top + Kinetic energy at the top = Kinetic energy at the bottom.
mgh1 + (1/2)mv^2 = (1/2)mv^2'

Where:
v' is the speed of the cart at the bottom of the hill

Rearranging the equation, we can solve for v':

v'^2 = 2gh1 + v^2

Substituting the given values (v = 6 m/s), we get:

v'^2 = 2 * 9.8 * H1 + 6^2

v' = sqrt(2 * 9.8 * H1 + 36)

Therefore, the speed of the cart at the bottom of the hill, in both scenarios, is given by the equations:

Scenario 1: v = sqrt(2 * 9.8 * H1)
Scenario 2: v' = sqrt(2 * 9.8 * H1 + 36)