A 10 kg block slides up a hill to a height of 5.8 m. If 414 J of thermal energy are generated, how fast was the block going at the bottom of the hill?
1 m/s
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To find the speed of the block at the bottom of the hill, we can use the principle of conservation of energy. The initial potential energy of the block is converted into kinetic energy at the bottom of the hill.
First, let's calculate the potential energy of the block at the top of the hill using the formula:
Potential Energy = mass * gravity * height
where mass = 10 kg, gravity = 9.8 m/s^2 (approximate value), and height = 5.8 m.
Potential Energy = 10 kg * 9.8 m/s^2 * 5.8 m = 568.4 J (rounded to one decimal place)
Now, let's calculate the kinetic energy of the block at the bottom of the hill using the formula:
Kinetic Energy = 568.4 J - 414 J = 154.4 J (rounded to one decimal place), considering the thermal energy generated.
Lastly, we can find the velocity (speed) of the block at the bottom of the hill using the formula:
Kinetic Energy = 0.5 * mass * velocity^2
Rearranging the equation, we have:
velocity^2 = (2 * Kinetic Energy) / mass
Substituting the values, we get:
velocity^2 = (2 * 154.4 J) / 10 kg = 30.88 m^2/s^2 (rounded to two decimal places)
Taking the square root of both sides, we find:
velocity ≈ √(30.88 m^2/s^2) ≈ 5.55 m/s (rounded to two decimal places)
Therefore, the speed of the block at the bottom of the hill is approximately 5.55 m/s.