(A) The 4-kg mass travels twice as far as the 2-kg mass before stopping. (B) The 2-kg mass travels twice as far as the 4-kg mass before stopping. (C) Both masses travel the same distance before stopping.

(D) The 2 kg mass travels greater than twice as far.

(E) None of the above

The correct answer is (B) The 2-kg mass travels twice as far as the 4 kg mass before stopping. To calculate the distance the object will slide one can use the work-energy theorem that states that the work done to a system is equal to the system's change in energy.

Here, the object starts with kinetic energy [KE = (mv^2)/2] and friction brings it to rest by doing an equivalent amount of work. Work is, in this case, the product of the force of friction and the distance the object slides while coming to rest. Therefore, the distance the object slides is directly proportional to the kinetic energy of the object, and inversely proportional to the force of friction, i.e. d = (mv^2)/(2F).

The 2-kg mass is half the mass of the 4-kg mass, and from this alone would have half of the kinetic energy as well. However, in this case the 2-kg mass is moving twice as fast as the 4-kg mass, and because its kinetic energy is dependent on the square of the velocity it turns out that the 2-kg mass has twice the kinetic energy of the 4-kg mass and will therefore slide twice as far before stopping.

2)A bird is standing in a bird cage, which is itself resting on an electronic balance. The bird flaps its wings and starts to fly upwards while remaining in the cage. Neglecting any effects from the turbulence of the air created by the flying bird, will the mass reading of the...

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