A 4.0kg block with an initial speed of 5.0 m/s slides up a rough inclined plane

Question

A 4.0kg block with an initial speed of 5.0 m/s slides up a rough inclined plane (which makes an angle of 30 degrees with the horizontal) and travels a distance of 2.0m (measured along the incline) before stopping.

1. By how much does the kinetic energy of the block change?

2. By how much does the potential energy of the system change?

3. By how much does the internal energy of the system change?

4. What is the magnitude of the constant frictional force?

5. If the incline was frictionless, but everything else was the same, how far up the incline would the block travel?

Thank you very much for helping.

Explanation / Answer

1)1/2 m*v^2

=50J

2)mgh

=2*9.8*2*sin30

=19.6 J

3)chANge in internal energy

=change in total energy

=50-19.6=29.4 J

4)F*2=29.4

F=14.7 N

5)(v^2/2g)/sin30

=2.55 m up the inclined plane

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