Question

Calculate the potential energy, kinetic energy, mechanical energy, velocity,

and height of the skater at the various locations.

Answer 1

`Point 1, PE = 0 J, KE=1920 J, ME=1920JPoint 2, PE = 588 J, KE= 1331 J, ME= 1919 J, v=6.66 m/sPoint 3, PE = 1923 J, KE=0 J ,ME= 1923 J, v=0 m/s, h=3.27 m`

How to determine the the potential energy, kinetic energy, mechanical energy, velocity?

Given Data:

Mass (m) = 60 kg

At Point 1:
`\( v_1 = 8 \, \text{m/s} \)`

At Point 2:
`\( y_2 = 1 \, \text{m} \)`

At Point 3:
`\( v_3 = 0 \, \text{m/s} \)`

Equations:

Kinetic Energy (KE):
`\( K = (1)/(2)mv^2 \)`

Gravitational Potential Energy (Ug):
`\( U_g = mgy \)`

Mechanical Energy (EMech):
`\( E_{\text{Mech}} = U_g + K \)`

Conservation of Energy (E₁ = E₂ = E₃): Energy is conserved throughout the system.

Point 1:

`\( U_(g1) = 0 \, \text{J} \)` (at ground level)

`\( K_1 = (1)/(2) * 60 * 8^2 = 1920 \, \text{J} \)`

`\( E_{\text{Mech}_1} = U_(g1) + K_1 = 1920 \, \text{J} \)`

Point 2:

`\( U_(g2) = mgy_2 = 60 * 9.8 * 1 = 588 \, \text{J} \)`

Using Conservation of Energy:
`\( U_(g2) + K_2 = 1920 \, \text{J} \)\( K_2 = 1920 - 588 = 1332 \, \text{J} \)`

Find velocity at Point 2:
`\( K_2 = (1)/(2) * 60 * v_2^2 \)`

`\( v_2 = \sqrt{(1332)/(30)} \approx 6.66 \, \text{m/s} \)`

`\( E_{\text{Mech}_2} = U_(g2) + K_2 = 588 + 1332 = 1920 \, \text{J} \)`(remains constant)

Point 3:

Using Conservation of Energy between Points 1 and 3:
`\( U_(g1) = U_(g3) \)`

`\( U_(g3) = mgh = 60 * 9.8 * h = 1920 \, \text{J} \)`

`As \( v_3 = 0 \, \text{m/s} \)` at Point 3, all energy is potential.

`\( h = (1920)/(60 * 9.8) = 3.27 \, \text{m} \)`

`\( U_(g3) = 60 * 9.8 * 3.27 = 1923 \, \text{J} \)`

Remember, at each point, the mechanical energy (sum of kinetic and potential energies) remains constant due to the conservation of energy principle.