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Sergio [31]
2 years ago
9

A 13,000 N car starts at rest and rolls down a hill froma height of 10 m. It then moves across a level surfaceand collides with

a light spring-loaded guardrail. Neglecting any losses due to friction, find the maximum distance the spring is compressed. Assume a spring constant of 1.0 x106N/m.
Physics
1 answer:
Leto [7]2 years ago
6 0

Answer:

The spring is compressed 0.51 m.

Explanation:

Hi there!

To solve this problem, we have to use the theorem of the conservation of energy that states that energy can´t be created nor destroyed, it can only be transformed.

In the case of the car, it has initially only gravitational potential energy because it is at rest and at a height of 10 m. The potential energy (PE) is calculated as follows:

PE = m · g · h

Where:

m · g = weight of the car

h = height at which the car is located

As the car rolls down the hill, the potential energy decreases because the height at which the car is located decreases. Since energy can´t be lost, this decrease in potential energy causes an increase of equal magnitude of the kinetic energy of the car. When the car is at the bottom of the hill, all the initial potential energy will have converted into kinetic energy (KE):

PE at the top = KE at the bottom

(notice that this is only possible because there is no friction, in which case some energy would be converted into thermal energy and the PE at the top would be greater than the KE at the bottom).

Then the car travels across a level surface and collides with a spring. As the car begins to stop, the kinetic energy decreases. In this case, the increasing energy is the elastic potential energy (EPE). When the car finally stops, all the kinetic energy of the car at the bottom of the hill will have transformed into elastic potential energy. Then:

KE at the bottom = EPE

The elastic potential energy is calculated as follows:

EPE = 1/2 · k · x²

Where:

k = spring constant.

x = compressing distance.

Since:

PE at the top = KE at the bottom

and

KE at the bottom = EPE

Then:

PE at the top = EPE

m · g · h = 1/2 · k · x²

Solving for x:

2 · m · g · h /  k = x²

x = √(2 · m · g · h /  k )

x = √(2 · 13000 N · 10 m / 1.0 × 10⁶ N/m)

x = 0.51 m

The spring is compressed 0.51 m.

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Assuming the Earth's mass but not its diameter increased, in the equation above m1 (the term usually indicative of the object of larger mass) would increase, while the r^2 would not.

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