Law of Conservation of Energy Problems with Solutions

Problem 1:

An intrepid physics student decides to try bungee jumping. She obtains a cord that is h=12 m long and has a spring constant of K=3.00 * 10^2 N/m. When fully suited, she has a mass of m=69 kg. She looks for a bridge to which she can tie the cord and step off. Determine the minimum height of the bridge L, that will allow her to stay dry (that is, so that she stops just before hitting the water below). Assume air resistance that is negligible.

law-conservation-of-energy-problem

Solution:

From the conservation of energy, we have kinetic and elastic energies that have transformed into potential energy:

\frac{mv^2}{2} + \frac{K(L-h)^2}{2} = mgL

Where

\frac{mv^2}{2} = mgh

Therefore

mgh = \frac{K(L-h)^2}{2} = mgL

K*L^2 - 2K*L*h + K*h^2 - 2mgL = 0

K*L^2 - 2(K*h+mg)L + K*h^2 = 0

3*10^2 * L^2 - 2(3*10^2*12 + 69*9.81)L + 3*10^2*12^2 = 0

3*10^2*L^2 - 8553.78*L + 43200 = 0

We have two solutions:

L = 21.95 and L = 6.56

The height of the bridge can’t be less than length of cord, therefore we select only L = 21.95 m.

Answer: L = 21.95 m.

Problem 2:

Tony Hawk (m=66kg) rides his skateboard at a local skate park. He starts from rest at the top of the track as seen in the figure below and begins a descent down the track, always maintaining contact with the surface. The mass of the skateboard is negligible, as is friction except where noted.

law-conservation-of-energy-problem-with-solutions-2

(a) What is Shawn’s speed when he reaches the bottom of the initial dip, 18.0 m below the starting point?

(b) He then ascends the other side of the dip to the top of a hill, 8.0 m above the ground. What is his speed when he reaches this point?

Solution:

a. From the conservation of energy: Potential energy at the top of the 18 m transforms into the Kinetic energy at the bottom of the dip.

mgh_1 = \frac{mv^2_1}{2}

v_1 = \sqrt{2gh_1} = \sqrt{2*9.81*18} = 18.8_{m/s}

b. From the conservation of energy: Potential energy at the top of the 18 m transforms into the Kinetic and Potential energy at the top of a hill.

mgh_1 = \frac{mv^2_2}{2} + mgh_2

mg(h_1 - h_2) = \frac{mv^2_2}{2}

v_2 = \sqrt{2g(h_1-h_2)} = \sqrt{2*9.81*(18-10)} = 12.5_{m/s}

Answer v_1 = 18.8_{m/s} and v_2 = 12.5_{m/s}

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