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Answer to Question #119512 in Mechanics | Relativity for vasenai

Question #119512

a) Johnis riding a bicycle and comesacrossa hill of height7.30 m. At the base of the hill, he is traveling at 6.00 m/s. When he reaches the top of the hill, he is traveling at1.00 m/s. Jonathan and his bicycle together have a mass of 85.0 kg. Ignore friction in the bicycle mechanism and between the bicycle tires and the road. i)What is the total external work done on the system of Jonathan and the bicycle between the time he starts up the hill and the time he reaches the top? ii) What is the change inpotential energy stored in Jonathan’s body during this process?

1
2020-06-01T14:24:08-0400

Explanations & Calculations

• If he let his initial kinetic energy to be exhausted into potential energy only, he will stop by a height about 1.8 m from the base.

"\\qquad\\qquad\n\\begin{aligned}\n\\small E_{k1} + E_{P1} & = E_{k2} + E_{P2}\\\\\n\\small \\frac{1}{2}\\times 85kg \\times (6 ms^{-1})^2 &= \\small85kg \\times9.8\\times h\\\\\n\\small h &= \\small 1.837m \n\\end{aligned}"

• Therefore, to make it to the top along with a velocity, he needs to do some extra work on the system.
• It's the difference between the final & the initial energies he possesses.

1) Work,

"\\qquad\\qquad\n\\begin{aligned}\n&= \\small E_f - E_i\\\\\n&= \\small\\Big(mgH + \\frac{1}{2}mv_2^2 \\Big) - \\frac{1}{2}mv_1^2\\\\\n&= \\small mgH +\\frac{1}{2}m(v_2^2-v_1^2)\\\\\n&= \\small 85\\times9.8\\times7.3m +\\frac{1}{2}\\times85\\times(1^2-6^2)\\\\\n&= \\small \\bold{4593.4 J}\n\\end{aligned}"

2) Potential energy difference,

"\\qquad\\qquad\n\\begin{aligned}\n\\small E_{pf} -E_{pi} &= \\small mgH -mgh \\\\\n&= \\small mgH -0 \\cdots{(\\because h = 0 \\text{as it's the base})}\\\\\n& =\\small 85kg \\times 9.8\\times 7.3m\\\\\n&= \\small \\bold{6080.9 J}\n\n\\end{aligned}"

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