The diagram shows a 0.74-kg square block at four positions along a track. The tr

Question

The diagram shows a 0.74-kg square block at four positions along a track. The track is frictionless between points A and B. Elsewhere, the coefficient of static friction on the block is 0.36.

Initially, the block is held in place against a spring by a horizontal external force of magnitude 29 N. Then the external force is removed, and the block is projected forward with speed v(1)=1.2 m/s. The block descends a ramp and has speed v(2)=1.9 m/s at the bottom. At point C, its speed is v(3)=1.4 m/s. The block stops at point D.

What is the height h of the ramp?

2.6 m

34 cm

1.5 m

82 cm

11 cm

The bob of a pendulum swings back and forth with a total mechanical energy of 300 J. What is the kinetic energy of the bob when it is at the lowest point of its trajectory? Assume that friction does not affect the pendulum.

150 J

0 J

300 J

200 J

2.6 m

34 cm

1.5 m

82 cm

11 cm

Explanation / Answer

Hi,

In the first problem the diagram would have been of great help, but since it's not there I think I can imagine the situation.

This problem can be solved using the conservation of the mechanical energy. Once the block has left the spring, it does not posses any elastic potential energy, but since it is moving with certain speed and it is over a ramp, it has kinetic and potential gravitational energy.

Then, once the block has descended the ramp, it only has kinetic energy and if there is no friction while the block is moving from those points, then the mechanical energy should be conserved. Therefore:

E1 = E2 ::::::: mgh + (0.5) mv12 = (0.5) mv22 :::::::: h = (0.5)(v22 - v12 )/g

h = (0.5)( 1.92 - 1.22 )/(9.8) m = 0.11 m ::::::: h = 11 cm

So the answer should be the last one, 11 cm.

Note: the answer given depends a lot on my interpretation of the data. The diagram would have help in clearing any misunderstanding.

In the case of the second problem we almost have a theoric question. The mechanical energy of the bob of a pendulum is the combination of both: the gravitational potential energy and the kinetic energy. If the bod is at the lowest point of its trajectory, then the gravitational potential energy is zero (because there is not height to take into account) and the mechanical energy is equal to the kinetic one, therefore:

K = 300 J

I hope it helps.

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