Answer:
The speed of the block when it is 7 m above the ground is approximately 19.8 m/s.
Step-by-step explanation:
To find the speed of the block when it is h2 = 7 m above the ground, we can use the principle of conservation of energy.
The initial potential energy of the block at height h1 is given by:
PE1 = m * g * h1 ,
where m is the mass of the block (20 kg), g is the acceleration due to gravity (9.80 m/s^2), and h1 is the initial height (27 m).
The initial kinetic energy of the block is zero since it is released from rest.
At height h2 = 7 m, the potential energy of the block is:
PE2 = m * g * h2
The final kinetic energy of the block is:
KE2 = (1/2) * m * v2^2 ,
where v2 is the final velocity or speed we want to find.
According to the conservation of energy, the initial potential energy plus the initial kinetic energy is equal to the final potential energy plus the final kinetic energy:
PE1 + KE1 = PE2 + KE2
Since the initial kinetic energy (KE1) is zero, the equation simplifies to:
PE1 = PE2 + KE2
Substituting the expressions for potential energy and solving for v2:
m * g * h1 = m * g * h2 + (1/2) * m * v2^2
Canceling the mass (m) on both sides of the equation:
g * h1 = g * h2 + (1/2) * v2^2
Rearranging the equation to solve for v2:
v2^2 = 2 * g * (h1 - h2)
Taking the square root of both sides to find v2:
v2 = sqrt(2 * g * (h1 - h2))
Now we can substitute the given values:
g = 9.80 m/s^2
h1 = 27 m
h2 = 7 m
Calculating v2:
v2 = sqrt(2 * 9.80 * (27 - 7))
= sqrt(2 * 9.80 * 20)
= sqrt(392)
≈ 19.8 m/s