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.
