The cart starts from rest at the top of a hill with height, H2. It rolls down the hill and at the bottom of the hill it has a speed of 10 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 4.59 m/s. How fast is it going at the bottom of the hill?
To find the speed of the cart at the bottom of the hill in both scenarios, we can use the principle of conservation of energy. The initial potential energy at the top of the hill is converted into kinetic energy at the bottom of the hill.
We can use the following equation to calculate the speed at the bottom of the hill:
(1/2)m*v^2 = m*g*h
Where:
m = mass of the cart
v = velocity at the bottom of the hill
g = acceleration due to gravity (approximately 9.8 m/s^2)
h = height of the hill
Let's consider the first scenario, where the cart starts from rest at the top of the hill with height H2. We need to find the value of H2 to proceed.
Now, let's consider the second scenario, where the cart is pushed so that it starts at the top of the same hill with a speed of 4.59 m/s. We need to find the speed of the cart at the bottom of the hill.
Using the conservation of energy principle for the second scenario, we can write:
(1/2)m*v1^2 = m*g*h
Where:
v1 = initial velocity at the top of the hill in the second scenario
h = height of the hill
We are given v1 = 4.59 m/s, and we need to find the value of h in order to calculate the speed at the bottom of the hill.
Once we have the value of h for both scenarios, we can plug it into the equation (1/2)m*v^2 = m*g*h to find the speed at the bottom of the hill in each scenario.
Please provide the height of the hill (H2) to proceed further with the calculations.
Vf^2 = Vo^2 + 2g*h2.
Vf^2 =(4.59)^2 + 19.6*h2,
Vf^2 = 21.07 + 19.6h2,
Vf = sqrt(21.07 + 19.6h2).