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A 100-g toy car is propelled by a compressed spring that starts it moving. The car follows the curved track in Figure 7.39. Show that the final speed of the toy car is 0.687 m/s if its initial speed is 2.00 m/s and it coasts up the frictionless slope, gaining 0.180 m in altitude.

Short Answer

Expert verified

Proved that the final speed of the toy car is 0.687 m/s.

Step by step solution

01

Conservation of energy

Conservation of energy: The universe's total energy is conserved. According to conservation of energy, energy neither be created not be destroyed; only the form of energy can be changed.

For a mechanical system, the conservation of energy is given as,

KEi+PEi=KEf+PEf

Here, KEiis the initial kinetic energy, PEiis the initial potential energy, KEfis the final kinetic energy, and PEfis the final potential energy.

02

The Final speed of the toy car

From the conservation of energy,

KEi+PEi=KEf+PEf12mvi2+mghi=12mvf2+mghf (1.1)

Here, m is the mass of the toy car m=100g,vi is the initial speed of the toy car vi=2.00m/s, g is the acceleration due to gravity g=9.8m/s2,vf is the final speed of the toy car, and hfis the maximum height attained by the toy car on a frictionless slope hf=0.180m.

Rearranging equation (1.1) in order to get an expression for the final speed of the toy car,

vf=vi2+2ghi-hf

Putting all known values,

vf=2.00m/s2+2×9.8m/s2×0-0.180m=0.687m/s

Hence proved, the final speed of the toy car is 0.687m/s.

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Most popular questions from this chapter

Suppose a car travels\(108{\rm{ km}}\)at a speed of\(30.0\,{\rm{m}}/{\rm{s}}\), and uses\(2.0{\rm{ gal}}\)of gasoline. Only\(30\% \)30% of the gasoline goes into useful work by the force that keeps the car moving at constant speed despite friction. (See Table 7.1 for the energy content of gasoline.)

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