A roller coster cart has a mass of 338kg.?

Question

The cart starts from a point A, which is stationary, and has a height of 24.5m. A while later the cart reaches a point B, which has a height 11.9m. Acceleration of gravity is 9.8m/s^2. What is the potential energy of the cart relatiove to the ground at point A (in units of J)? Also What is the speed of the cart at point B, ignore friction(in m/s)?

Answer 1

To solve for the potential energy @ A, use:

P.E. = m*g*height = (338 kg)(9.8m/s^2)*(24.5m) = 8.12x10^4 J

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To find the speed of the cart @ B, do the following steps:

We know that @ B, both kinetic and potential energy are present. Since we are ignoring friction, we assume that energy is conserved.

Thus, energy @ A and B are the same.
This means P.E(@ B) +K.E.(@B) = P.E.(@A) = 8.12x10^4 J

We can solve for the potential energy @ B:
P.E. = m*g*height at B = (338 kg)*(9.8m/s^2)*(11.9 m)
= 3.94x10^4 J

Therefore, the kinetic energy @ B is P.E.(@A)-P.E.(@B) = 8.12X10^4-3.94X10^4 = 4.18X10^4 J

Using the kinetic energy @ B, we can solve for the speed of the cart:

K.E.(@B) = 4.18x10^4 = 0.5*m*v^2, solving for 'v', you get:

v = sqrt(2*K.E./m) = sqrt(2*4.18x10^4/338) = 247.34 m/s

Hope this helps

Answer 2

delta P.E. is roughly g*m*delta height.

K.E. is about 0.5*m*v^2.

since m are the same...g*delta height = 0.5 * v^2

v = sqrt ( 2 * g * delta height )

Answer 3

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