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?

Initial kinetic energy = (1/2) M Vo^2

= (thermal energy generated) + (potential energy increase)
= 414 J + M g H = 982 J

Solve for Vo

To find the speed of the block at the bottom of the hill, we can use the principle of conservation of mechanical energy.

The mechanical energy of the block is the sum of its kinetic energy (KE) and potential energy (PE):

ME = KE + PE

At the top of the hill, all of the block's energy is in the form of potential energy, given by:

PE = m * g * h

where m is the mass of the block (10 kg), g is the acceleration due to gravity (9.8 m/s²), and h is the height of the hill (5.8 m).

PE = 10 kg * 9.8 m/s² * 5.8 m
= 568.4 Joules

At the bottom of the hill, all of the block's energy is in the form of kinetic energy. Therefore, we can equate the potential energy at the top to the kinetic energy at the bottom:

KE = 568.4 J

The kinetic energy is given by:

KE = 0.5 * m * v²

where v is the velocity/speed of the block at the bottom of the hill. Rearranging this equation, we can solve for v:

v² = 2 * KE / m

v² = 2 * 568.4 J / 10 kg
v² = 113.68 m²/s²

Taking the square root of both sides, we find:

v ≈ 10.67 m/s

Therefore, the block was going approximately 10.67 m/s at the bottom of the hill.