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leva [86]
3 years ago
14

To test the resiliency of its bumper during low-speed collisions, a 2 840-kg automobile is driven into a brick wall. The car's b

umper behaves like a spring with a force constant 6.00 106 N/m and compresses 3.98 cm as the car is brought to rest. What was the speed of the car before impact, assuming no mechanical energy is transformed or transferred away during impact with the wall
Physics
1 answer:
Westkost [7]3 years ago
4 0

Answer:

V= 1.82 m/s

Explanation:

Given that

mass , m = 2840 kg

Spring constant ,K = 6 x 10⁶ N/m

Compression in the spring , x = 3.98 cm

Lets take speed of the car before impact = V

Now by using energy conservation

\dfrac{1}{2}Kx^2=\dfrac{1}{2}mV^2

Kx^2=mv^2

V=\sqrt{\dfrac{Kx^2}{m}}

Now by putting the values in the above equation

V=\sqrt{\dfrac{6\times 10^6\times (3.98\times 10^{-2})^2}{2840}}

V= 1.82 m/s

Therefore the speed of the car before impact will be 1.82 m/s.

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A) The answer is 11.53 m/s

The final kinetic energy (KEf) is the sum of initial kinetic energy (KEi) and initial potential energy (PEi).
KEf = KEi + PEi

Kinetic energy depends on mass (m) and velocity (v)
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KEi = 1/2 m * vi²

Potential energy depends on mass (m), acceleration (a), and height (h):
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So:
KEf = KEi + <span>PEi
</span>1/2 m * vf² =  1/2 m * vi² + m * a * h
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We know:
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b) The answer is 6.78 m

The kinetic energy at the bottom (KE) is equal to the potential energy at the highest point (PE)
KE = PE

Kinetic energy depends on mass (m) and velocity (v)
KE = 1/2 m * v²

Potential energy depends on mass (m), acceleration (a), and height (h):
PE = m * a * h

KE = PE
1/2 m * v² = m * a * h

Divide both sides by m:
1/2 * v² = a * h
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a = 9.8 m/s² 
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1/2 * 11.53² = 9.8 * h
1/2 * 132.94 = 9.8 * h
66.47 = 9.8 * h
h = 66.47 / 9.8
h = 6.78 m
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