Question

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
I just not sure how I go about getting the answer.

Answer 1

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,

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

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

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

$\Rightarrow u=\sqrt{54.71}$

$u=7.4 m/sec$