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# Answer to Question #108647 in Mechanics | Relativity for Joey

Question #108647
A 62.7 kg skateboarder started at a height of 4.55 metres above the ground and finished at 5.70 metres above the ground. If the skateboarder ends up travelling at 6.00 m/s, and friction removes 120.0 J of energy, find the skateboarder’s original speed.
KE1 + PE1 = KE2 + PE2
1
2020-04-08T10:59:16-0400

As per the given question,

Mass of the stakeholder (m)=62.7 kg

Initial height "(h_1)=4.55 m"

Final height "(h_2)=5.7 m"

Final speed of the stakeholder (v)=6 m/sec

Work done by the friction =120J

Let the initial speed of the stakeholder is (u),

Now applying the conservation of energy,

"mgh_1+ \\dfrac{mu^2}{2}+W_{fs}=mgh_2+\\dfrac{mv^2}{2}"

Now substituting the values,

"62.7\\times 9.8\\times 4.55+ \\dfrac{62.7\\times u^2}{2}+120=62.7\\times 9.8\\times 5.7+\\dfrac{62.7\\times6^2}{2}"

"\\Rightarrow \\dfrac{62.7\\times u^2}{2}=3502.422+1128.6-120-2795.793"

"\\Rightarrow u^2=\\dfrac{1715.229\\times2}{62.7}"

"\\Rightarrow u=\\sqrt{54.71}"

"u=7.4 m\/sec"

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