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The mass of a bicyclist and a bicycle together is 53.0 kg. How much work has been
done if the bicyclist slows the bicycle from a speed of 3.84 m/s to 1.27 m/s?
(A) -68.1 J
(B) -136 J
(C) -348 J
(D) -696 J

Sagot :

Answer:

Approximately [tex](-348\; {\rm J})[/tex], assuming that the bike was on level ground.

Explanation:

If an object of mass [tex]m[/tex] is moving at a speed of [tex]v[/tex], the kinetic energy of that object would be [tex](1/2)\, m\, v^{2}[/tex].

At a speed of [tex]v = 3.84\; {\rm m\cdot s^{-1}}[/tex], the kinetic energy of the bike and the cyclist- combined- would be:

[tex]\begin{aligned} &\frac{1}{2}\, m\, v^{2} \\=\; &\frac{1}{2}\times 53.0\; {\rm kg} \times (3.84\; {\rm m\cdot s^{-2}})^{2} \\ \approx \; & 390.76\; {\rm J}\end{aligned}[/tex].

At a speed of [tex]v = 1.27\; {\rm m\cdot s^{-1}}[/tex], the kinetic energy of the bike and the cyclist- combined- would be:

[tex]\begin{aligned} &\frac{1}{2}\, m\, v^{2} \\=\; &\frac{1}{2}\times 53.0\; {\rm kg} \times (1.27\; {\rm m\cdot s^{-2}})^{2} \\ \approx \; &42.741 \; {\rm J}\end{aligned}[/tex].

The energy change was approximately:

[tex]390.76\; {\rm J} - 42.741\; {\rm J} \approx 348\; {\rm J}[/tex].

If the bike was on level ground, the friction on the bike and the cyclist would have done a work of [tex](-348\: {\rm J})[/tex] to reduce the kinetic energy of the bike and the cyclist by that amount.

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