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3 years ago

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A 100-g toy car moves along a curved frictionless track. At first, the car runs along a flat horizontal segment with an initial

velocity of 2.77 m/s. The car then runs up the frictionless slope, gaining 0.184 m in altitude before leveling out to another horizontal segment at the higher level. What is the final velocity of the car if we neglect air resistance?

1 answer:

aleksandrvk [35]3 years ago

7 0

Answer:

The final velocity of the car is 2.02 m/s

Explanation:

Hi there!

The kinetic energy of the car as it runs along the first flat horizontal segment can be calculated using the following equation:

KE = 1/2 · m · v²

Where:

KE = kinetic energy

m = mass

v = velocity

Then, the initial kinetic energy will be:

KE = 1/2 · 0.100 kg · (2.77 m/s)²

KE = 0.384 J

When the car gains altitude, it gains potential energy. The amount of gained potential energy will be equal to the loss of kinetic energy. So let´s calculate the potential energy of the car as it reaches the top:

PE = m · g · h

Where:

PE = potential energy.

m = mass

g = acceleration due to gravity.

h = height.

PE = 0.100 kg · 9.8 m/s² · 0.184 m

PE = 0.180 J

Then, the final kinetic energy will be (0.384 J - 0.180 J) 0.204 J

Using the equation of kinetice energy, we can obtain the velocity of the car:

KE = 1/2 · m · v²

0.204 J = 1/2 · 0.100 kg · v²

2 · 0.204 J / 0.100 kg = v²

v = 2.02 m/s

The final velocity of the car is 2.02 m/s

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