A car of mass 1100.0 kg is at the top of a 30.00 degree hill. The distance d is

Question

A car of mass 1100.0 kg is at the top of a 30.00 degree hill. The distance d is 28.00 m. The car begins at rest. Neglect fiction. a. What is its gravitational potential energy relative to the bottom of the hill? b. If the car rolls down the hill (with the engine off) with negligible friction and air resistance, what will its kinetic energy be when it reaches the bottom? Explain c. What will its speed be when it reaches the bottom? d. The car (with that speed) now rolls up a (still frictionless) hill with a steeper slope. How high (vertical distance) above the bottom of the hill does it get before coming to a (momentary) stop?

Explanation / Answer

a) mgh = 1100*9.8*14 = 150920J b) K.e = 150920J by the law of conservation of energy 1/2mv^2 = 150920 v = 16.565

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