Question

A 2kg block has 70J of KE. It then travels 1.5 meters up a hill. As it travels up the hill friction does -12J of work on the block. What is the new speed of the block?
Please include an explanation.

Answer 1

Answer:

v = 5.34[m/s]

Explanation:

In order to solve this problem, we must use the theorem of work and energy conservation. This theorem tells us that the sum of the mechanical energy in the initial state plus the work on or performed by a body must be equal to the mechanical energy in the final state.

Mechanical energy is defined as the sum of energies, kinetic, potential, and elastic.

E₁ = mechanical energy at initial state [J]

$E_{1}=E_{pot}+E_{kin}+E_{elas}$

In the initial state, we only have kinetic energy, potential energy is not had since the reference point is taken below 1.5[m], and the reference point is taken as potential energy equal to zero.

In the final state, you have kinetic energy and potential since the car has climbed 1.5[m] of the hill. Elastic energy is not available since there are no springs.

E₂ = mechanical energy at final state [J]

$E_{2}=E_{kin}+E_{pot}$

Now we can use the first statement to get the first equation:

$E_{1}+W_{1-2}=E_{2}$

where:

W₁₋₂ = work from the state 1 to 2.

$E_{k}=\frac{1}{2} *m*v^{2}$

$E_{pot}=m*g*h$

where:

h = elevation = 1.5 [m]

g = gravity acceleration = 9.81 [m/s²]

$70 - 12 = \frac{1}{2}*2*v^{2}+2*9.81*1.5$

$58 = v^{2} +29.43\\v^{2} =28.57\\v=\sqrt{28.57}\\v=5.34[m/s]$