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
Potential energy at the top of the hill
= mgh
= 10*9.81*5.8 J
=569 J
Heat energy produced
= 414 J
If we assume conservation of energies, energy at the bottom of the hill in the form of kinetic energy
=(1/2)mv² = 569+414 =983
Solve for v:
v=sqrt(2*983/10)
=14.0 m/s
To determine the speed of the block at the bottom of the hill, we need to apply the principle of conservation of energy.
First, let's calculate the potential energy (PE) gained by the block as it moves up the hill. The formula for potential energy is given by PE = m * g * h, where m is the mass, g is the acceleration due to gravity (9.8 m/s^2), and h is the height. Substituting the given values, we have PE = 10 kg * 9.8 m/s^2 * 5.8 m = 568 J.
Now, we know that the total energy at the bottom of the hill is equal to the sum of the potential energy gained and the thermal energy generated. So, the total energy at the bottom of the hill is 568 J + 414 J = 982 J.
The total energy at the bottom of the hill is equal to the kinetic energy (KE) of the block. The formula for kinetic energy is given by KE = (1/2) * m * v^2, where v is the velocity of the object. Rearranging the formula, we can solve for the velocity as follows:
v^2 = (2 * KE) / m
Substituting the values, we have v^2 = (2 * 982 J) / 10 kg = 196.4.
Taking the square root of both sides, v = sqrt(196.4) ≈ 14 m/s.
Therefore, the block was traveling at approximately 14 m/s at the bottom of the hill, not 1 m/s.