Ablock of mass m1 = 3.5 kg moves with velocity v1 = 6.3 m/s on a frictionless surface. it collides with block of mass m2 = 1.7 kg which is initially stationary. the blocks stick together and encounter a rough surface. the blocks eventually come to a stop after traveling a distance d = 1.85 m . what is the coefficient of kinetic friction on the rough surface? μk =

The coefficient of kinetic friction is 0.49.

Explanation:

The mass of the block

Velocity

The mass of the block

Now, use the conservation of momentum principle:

we, can write:

on putting the values in above equation

we get:

Now the work done by the frictional force is:

Where , d is distance covered by the two blocks.

The initial kinetic energy

Now, the loss in kinetic energy must be equal to work done by friction force:

on putting the values in above equation:

Hence , the coefficient of kinetic friction is 0.49.

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total distance travelled = 60 miles

total time taken =1.5 hours

average speed = total distance / total time

60 / 1.5

= 40

therefore, average speed = 40 m/h

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answer would be letter choice (a),   a ball resting on a shelf. would be another example of potential energy   >   stored energy > not moving.

a ball spinning on a finger   >   kinetic energy > moving energy

a ball rolling on the ground   >   kinetic energy   > moving energy

a   ball flying through the air   > kinetic energy   >   moving energy

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