A metal ball falls from a height of 1m on a steel plate and jumps up to a height of 0.81m. Find the coefficient of restitution. A variation of a force in a certain region is given... A metal ball falls from a height of 1m on a steel plate and jumps up to a height of 0.81m. Find the coefficient of restitution. A variation of a force in a certain region is given by F = 6x^2 - 4x - 8. It displaces an object from x = 1m to x = 2m. Calculate the amount of work done. Read more

Steps to Solve

1. Calculate the Coefficient of Restitution

The coefficient of restitution (e) is defined as the ratio of the height after the bounce to the height before the bounce.

Using the formula:

$$ e = \sqrt{\frac{h'}{h}} $$

where $h'$ is the height after the bounce (0.81 m) and $h$ is the height before the bounce (1 m).

Thus,

$$ e = \sqrt{\frac{0.81}{1}} = \sqrt{0.81} = 0.9 $$

2. Find the Work Done by the Varying Force

To calculate the work done (W) by a varying force over a distance, use the formula:

$$ W = \int_{a}^{b} F(x) , dx $$

where $F(x) = 6x^2 - 4x - 8$, $a = 1$, and $b = 2$.

3. Set Up the Integral

First, set up the integral:

$$ W = \int_{1}^{2} (6x^2 - 4x - 8) , dx $$

4. Calculate the Integral

Now calculate the integral by finding the antiderivative:

$$ W = \left[ 2x^3 - 2x^2 - 8x \right]_{1}^{2} $$

5. Evaluate the Integral from 1 to 2

Substituting the bounds into the antiderivative:

$$ W = \left( 2(2)^3 - 2(2)^2 - 8(2) \right) - \left( 2(1)^3 - 2(1)^2 - 8(1) \right) $$

Calculating each term:

For $x = 2$:

$$ = 2(8) - 2(4) - 16 = 16 - 8 - 16 = -8 $$

For $x = 1$:

$$ = 2(1) - 2(1) - 8 = 2 - 2 - 8 = -8 $$

Now, substituting back into the equation for work done:

$$ W = -8 - (-8) = -8 + 8 = 0 $$