A child and sled with a combined mass of 59.0 kg slide down a frictionless slope. If the sled starts from rest and has a speed of 3.40 m/s at the bottom, what is the height of the hill?

Question

A child and sled with a combined mass of 59.0 kg slide down a frictionless slope. If the sled starts from rest and has a speed of 3.40 m/s at the bottom, what is the height of the hill?

Answer 1

To find the height of the hill, we can use the principle of conservation of energy, which states that the total energy of a system remains constant.

The total energy at the top of the hill is given by:
E_top = mgh

The total energy at the bottom of the hill is given by:
E_bottom = (1/2)mv^2

Since there is no loss of energy due to friction, the total energy at the top is equal to the total energy at the bottom:

mgh = (1/2)mv^2

We can cancel out the common factors of mass (m) and simplify the equation to find the height (h):

gh = (1/2)v^2

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

Let's plug in the given values:
v = 3.40 m/s (speed at the bottom)
g = 9.8 m/s^2 (acceleration due to gravity)

h = (1/2)(3.40^2) / 9.8

h = (1/2)(11.56) / 9.8

h = 5.78 / 9.8

h ≈ 0.59 m

Therefore, the height of the hill is approximately 0.59 meters.

Answer 2

To determine the height of the hill, we can use the principle of conservation of energy. The initial potential energy (mgh) of the child and the sled at the top of the hill is converted to kinetic energy (1/2mv^2) at the bottom of the hill, since there is no friction to consider.

First, let's calculate the initial potential energy (mgh) of the child and the sled at the top of the hill. We'll use the combined mass of the child and sled, which is 59.0 kg, and the acceleration due to gravity, which is approximately 9.8 m/s^2.

Potential energy = mass x gravity x height

mgh = (59.0 kg)(9.8 m/s^2)(h)

Next, let's calculate the final kinetic energy (1/2mv^2) at the bottom of the hill. We'll use the combined mass of the child and sled and the final velocity, which is 3.40 m/s.

Kinetic energy = 1/2 x mass x velocity^2

1/2mv^2 = 1/2(59.0 kg)(3.40 m/s)^2

Since energy is conserved, the initial potential energy is equal to the final kinetic energy. Therefore, we can set the equations equal to each other:

mgh = 1/2mv^2

(59.0 kg)(9.8 m/s^2)(h) = 1/2(59.0 kg)(3.40 m/s)^2

Now, we can solve for the height (h):

h = (1/2)(3.40 m/s)^2 / (9.8 m/s^2)

h = (0.5)(3.40 m/s)^2 / (9.8 m/s^2)

h = (0.5)(11.56 m^2/s^2) / (9.8 m/s^2)

h ≈ 0.591 m

Therefore, the height of the hill is approximately 0.591 meters.

Answer 3

mass not relevant

(1/2) m v^2 = m g h
mass cancels
v^2 = 2 g h
v = sqrt (2 g h)
or
h = v^2/(2 g)