Samuel drops a 2kg ball from some height above the surface of a table. The ball starts from rest and loses 36 Joules of gravitational potential energy before reaching the table’s surface. What is the speed of the ball just before it hits the table?
If needed, g=10ms2
just set
1/2 mv^2 = 36
and solve for v
To find the speed of the ball just before it hits the table, we can use the principle of conservation of energy. The initial potential energy of the ball when it is dropped is equal to the final kinetic energy just before it hits the table.
The potential energy of an object at height h is given by the formula: P.E = m * g * h, where m is the mass of the object, g is the acceleration due to gravity, and h is the height.
Given:
Initial potential energy (P.E) = 36 Joules
Mass of the ball (m) = 2 kg
Acceleration due to gravity (g) = 10 m/s^2
We can rearrange the equation for potential energy to solve for the height (h):
h = P.E / (m * g)
Substituting the given values:
h = 36 J / (2 kg * 10 m/s^2)
h = 1.8 m
Now, we can use the principle of conservation of energy to find the final speed of the ball just before hitting the table. The initial potential energy is converted into the final kinetic energy.
Final kinetic energy (K.E) = m * v^2 / 2, where v is the velocity (speed) of the ball just before it hits the table.
Since the ball starts from rest, the initial velocity (u) is zero. Therefore, the equation for conservation of energy becomes:
P.E = K.E
m * g * h = m * v^2 / 2
Substituting the values:
2 kg * 10 m/s^2 * 1.8 m = 2 kg * v^2 / 2
Simplifying the equation:
20 m/s^2 * 1.8 m = v^2
36 m^2/s^2 = v^2
Taking the square root of both sides:
v = √(36 m^2/s^2)
v = 6 m/s
Therefore, the speed of the ball just before it hits the table is 6 m/s.