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A roller coaster cart of mass m = 291 kg starts stationary at point A, where h1 = 35.2 m and a while later is at B, where h2 = 14.4 m. The acceleration of gravity is 9.8 m/s 2 . What is the potential energy of the cart relative to the ground at A? Answer in units of J.

Part 2: What is the speed of the cart at B, ignoring the effect of friction? Answer in units of m/s.
asked Jan 25, 2016 in PHYSICS by anonymous

1 Answer

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Step 1:

(1)

The initial velocity of the roller coaster at point A is 0 m/sec.

Mass of the roller coaster cart is image.

The height of mass at point A is .

The height of mass at point B is .

The acceleration due to gravity is 9.8 m/sec².

Find the potential eneregy at point A.

Potential energy at point A is .

image

Potential energy at point A is 100.38 kJ.

Step 2:

(2)

Find the speed of the cart at point B.

Equation of motion: image.

The acceleration due to freely falling body is equal to acceleration due to gravity.

image

The speed of the cart at point B is 20.19 m/sec.

Solution:

(1) Potential energy at point A is 100.38 kJ.

(2) The speed of the cart at point B is 20.19 m/sec.

answered Jan 26, 2016 by Lucy Mentor

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