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oksian1 [2.3K]
3 years ago
5

How does kinetic energy affect the stopping distance of a vehicle traveling at 30 mph compared to the same vehicle traveling at

60 mph?
Physics
1 answer:
Maru [420]3 years ago
5 0
Well you of course have different kinetic energies with the two speeds.
Kinetic energy = (1/2)*mass*velocity^2
The vehicle's mass is the same in both cases, so we can ignore that as well as 1/2 since it's a constant.
So we have (30)^2 vs (60^2)
which is 900 vs 3600
So having 60 mph compared to 30 mph is 4 times the kinetic energy.
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A block of ice(m = 14.0 kg) with an attached rope is at rest on a frictionless surface. You pull the block with a horizontal for
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a) The weight and the normal force of the block has a magnitude of 137.298 newtons and the pull force exerted on the block has a magnitude of 98 newtons.

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g  - Gravitational acceleration, measured in meters per square second.

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N = W = (14\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)

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F_{pull} = 98\,N

The weight and the normal force of the block has a magnitude of 137.298 newtons and the pull force exerted on the block has a magnitude of 98 newtons.

b) Since the block of ice is on a frictionless surface and pull force is parallel to the direction of motion and uniform in time, we can apply the Impact Theorem, which states that:

m\cdot v_{o} +\Sigma F \cdot \Delta t = m\cdot v_{f} (2)

Where:

v_{o}, v_{f} - Initial and final speeds of the block, measured in meters per second.

\Sigma F - Horizontal net force, measured in newtons.

\Delta t - Impact time, measured in seconds.

Now we clear the final speed in (2):

v_{f} = v_{o}+\frac{\Sigma F\cdot \Delta t}{m}

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v_{f} = 0\,\frac{m}{s}+\frac{(98\,N)\cdot (1.40\,s)}{14\,kg}

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The final speed of the block of ice is 9.8 meters per second.

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