A mass was dropped from a height of 20m. (a) find its speed at impact. (b) what can you say about the initial GPE of the mass and the KE of the mass just as it hits the ground?
To answer these questions, we can use the principles of conservation of energy. Here's how we can approach it:
(a) To find the speed at impact, we can use the principle of conservation of mechanical energy. At the top of the drop, all the energy is in the form of gravitational potential energy (GPE). At the bottom of the drop, all the energy is in the form of kinetic energy (KE). Therefore, we can equate the GPE at the top with the KE at the bottom:
GPE = KE
The gravitational potential energy is given by the formula:
GPE = m * g * h
Where:
m = mass of the object (unspecified in the question)
g = acceleration due to gravity (approximately 9.8 m/s^2)
h = height of the drop (20m)
The kinetic energy is given by the formula:
KE = 1/2 * m * v^2
Where:
v = velocity of the object at impact (what we need to find)
By equating the GPE with the KE, we get:
m * g * h = 1/2 * m * v^2
Simplifying the equation by canceling out the mass (m) from both sides, we are left with:
g * h = 1/2 * v^2
Rearranging the equation to solve for v, we get:
v^2 = 2 * g * h
Taking the square root of both sides gives us:
v = √(2 * g * h)
Plugging in the values of g (9.8 m/s^2) and h (20m) into the equation, we can calculate the speed at impact.
(b) As the mass hits the ground, its kinetic energy is at its maximum, and the gravitational potential energy is reduced to zero. This means that all the initial GPE is converted into KE just as it hits the ground. In other words, the initial GPE of the mass is equal to the KE of the mass at impact.
I hope this explanation helps you understand how to find the speed at impact and the relationship between the initial GPE and KE in this scenario.