A .10 kg ball dropped vertically from a height of 1 meter above the floor bounces back to a height of .80 m?

Question

the mechanical energy lost by the ball as it bounces is approximately

.30
.20
.78
.080

Answer 1

Hello #1,

The potential energy it had at 1 meter:
m*g*h = .10 kg * 9.8 m/s^2 * 1 m = .98 J

The potential energy it has at .80 meter:
m*g*h = .10 kg * 9.8 m/s^2 * .80 m = .78 J

So it lost about .20 J or about 20 %

Answer 2

There are 2 approaches you are able to make certain this issue. the first way is to apply uncomplicated guidelines of action: we've y = 4.ninety six for top, a = 9.80 one for acceleration, and V(initial) = 0, because the article grow to be held nonetheless previous to being released. we prefer to discover V(very last). V(very last)^2 = V(initial)^2 + 2a(y-y(initial)) y-y(initial) is basically the formula for displacement, and we calculate it as 0 - 4.ninety six, considering we prefer to subtract initial top from very last top. this supplies us -4.ninety six for displacement. We change our values in to get: V^2 = 0^2 + 2(-9.80 one)(-4.ninety six) make certain for V^2, then make certain for V. you are able to also use conservation of power to resolve this issue. ability power is completely switched over to kinetic power at the same time as the ball hits the floor, so: mgh = (a million/2)mv^2 (the following, our v is the speed at which the ball hits the floor, at the same time as all of ability power has been switched over to kinetic power) we may be able to simplify the equation and eliminate "m" to get: gh = (a million/2)v^2 (9.80 2)(4.ninety six) = (a million/2)v^2 make certain for v. As you are able to discover, conservation of power critically simplifies many "drop" complications.