In Fig 8.54, a block slides along a track from one level to a higher level after passing through an intermediate valley. The track is frictionless until the block reaches the higher level. There a frictional force stops the block in a distance d. The block’s initial speedv0is6.0m/s , the height difference his 1.1 m, andμkis0.60 . Find d.

Short Answer

Expert verified

The distance travelled by the block at the higher level is 1.2 m

Step by step solution

01

Listing the given quantities

The initial speed of the block = 6.0 m/s

The height difference between levels d = 1.1 m

The coefficient of friction is 0.60

02

Understanding the concept of conservation of energy

The block moving with initial speed travels past a frictionless intermediate valley. Then it rises to a higher level on the other side. Thus initial kinetic energy is converted to kinetic and potential energy. At this level, further track offers friction and kinetic energy of the block is lost as thermal energy and the block stops after travelling some distance. So, we apply the law of conservation of energy at each stage of motion.

Formula:

K.E=12mv2P.E=mghEth=Ffd

03

Calculation of distance travelled by the block at the higher level

Using the law of conservation of energy for the first frictionless part of motion, we determine the speed of the block as it reaches higher level

KEi=PEf+KEfKEF=KEI-PEF12mvf2=12mvf2=mgh12vf2=12×6.02-9.8×1.1vf2=2×18-10.8=14.4vf=3.8m/s

Now, this kinetic energy is lost as thermal energy. Hence we write

Eth=KEFμKmgd=12mvf2d=vf22μkg=14.42×0.60×9.8=1.2m

Hence, the distance travelled by the block at the higher level is 1.2 m

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