A ball of mass 0.5kg dropped from a cliff top,the ball hits the sea below at a speed of 10m/s
(a) what is the kinetic energy of the ball as it is about to hit the sea
(b) what was the balls gravitational potential energy before it was dropped
(c) from what height was the ball dropped
(a) The kinetic energy of an object is given by the equation:
KE = (1/2) * m * v^2
where KE is the kinetic energy, m is the mass of the object, and v is the velocity of the object. In this case, the mass of the ball is 0.5kg and the velocity is 10m/s. Plugging these values into the equation, we get:
KE = (1/2) * 0.5kg * (10m/s)^2
= 0.5 * 0.5kg * 100m^2/s^2
= 25 Joules
So, the kinetic energy of the ball as it is about to hit the sea is 25 Joules.
(b) The gravitational potential energy of an object at a certain height is given by the equation:
PE = m * g * h
where PE is the gravitational potential energy, m is the mass of the object, g is the acceleration due to gravity (which is approximately 9.81 m/s^2 on Earth), and h is the height. In this case, the mass of the ball is 0.5kg and the acceleration due to gravity is 9.81 m/s^2. We need to find the height at which the ball was dropped.
(c) Rearranging the equation for gravitational potential energy, we get:
h = PE / (m * g)
Plugging in the known values, we have:
h = (0.5kg * 9.81 m/s^2) / (0.5kg)
= 9.81 m
So, the ball was dropped from a height of 9.81 meters.
To solve this problem, we'll use the following equations:
(a) kinetic energy (K.E) = (1/2) * mass * velocity^2
(b) gravitational potential energy (G.P.E) = mass * gravitational acceleration * height
(c) velocity (v) = initial velocity + (acceleration * time)
(d) acceleration (a) = change in velocity / time
First, let's solve part (a) - the kinetic energy of the ball as it is about to hit the sea.
Given:
Mass (m) = 0.5 kg
Velocity (v) = 10 m/s
Using the formula for kinetic energy:
K.E = (1/2) * m * v^2
Substituting the given values:
K.E = (1/2) * 0.5 kg * (10 m/s)^2
K.E = 0.5 * 0.5 kg * 100 m^2/s^2 = 25 Joules
Therefore, the kinetic energy of the ball as it is about to hit the sea is 25 Joules.
Next, let's solve part (b) - the gravitational potential energy before the ball was dropped.
Given:
Mass (m) = 0.5 kg
The gravitational potential energy is equal to the work done against gravity when lifting the object to a certain height. The formula for gravitational potential energy is:
G.P.E = m * g * h
Where:
g = acceleration due to gravity
h = height
Substituting the given values:
G.P.E = 0.5 kg * 9.8 m/s^2 * h
Since the ball is being dropped, it starts with zero initial velocity. Therefore, all its initial energy is in the form of gravitational potential energy.
Since G.P.E is equal to K.E, we can equate them and solve for h:
25 J = 0.5 kg * 9.8 m/s^2 * h
Simplifying the equation:
25 J = 4.9 kg m^2/s^2 * h
h = 25 J / (4.9 kg m^2/s^2)
h ≈ 5.102 m
Therefore, the height from which the ball was dropped is approximately 5.102 meters.
In summary:
(a) The kinetic energy of the ball as it is about to hit the sea is 25 Joules.
(b) The ball's gravitational potential energy before it was dropped is approximately 5.102 meters.
(c) The ball was dropped from a height of approximately 5.102 meters.