These questions are connected A 2.0-kg mass is project from the edge of the top

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These questions are connected

A 2.0-kg mass is project from the edge of the top of a 20-m tall building with a velocity of 24 m/s at 30 degree above the horizontal. Disregard air resistance and assume the ground is level. What is the speed of the mass just before it strikes the ground? Show important physics steps of your calculations to earn full credits. 13 m/s 31 m/s 30 m/s 16 m/s 23 m/s none of the above If the mass in the problem is project at 60 degree above the horizontal instead, its speed just before it strikes the ground is more than less than equal to your answer in the above problem. A lightweight object and a very heavy object are sliding with equal speeds along a level frictionless surface. They both slide up the same frictionless hill. Which rises to a greater height? The heavy object, because it has greater kinetic energy. The lightweight object, because it weighs less. They both slide to the same height. cannot be determined from the information given In a given frictionless displacement of a particle, its kinetic energy increases by 25 J while its potential energy decreases by 10 J. Determine the work of the non-conservative forces acting on the particle during this displacement. -15 J -35 J +15 J -35 J +55 J none of the above

Explanation / Answer

part 1) here we will consider conservation of total energy principal

Initial potential energy= mgh

PEinitial=2 * 9.8 * 20=392 J

Initial kinetic energy= mu2/2

KEinitial=0.5 * 2 * 242=576 J

Final potential energy=0

Final kinetic energy=mv2/2

Total energy= constant

Total energy initial=total energy final

392 +576=0.5 * 2 * v2

v2=968

v=31.11 m/s so option B is the closest option

Part b) speed would be equal to the speed in the first case as we have seen in our previous part that it does not depend upon the angle, so option c is correct.

part c) here we willl again use conservation of energy

Kinetic energy will be converted to potential energy as they rise

so 1/2 m v2= mgh

h=v2/g

Here height h is independent of mass ,so both rise to the same height

option c is correct

Part d)workdone by non-conservative force =change in mechanical energy =10- 25= -15 J

so option a is correct

here - sign indicates that non-conservative force ( here friction) was acting opposite the direction of movement.

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