Question

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

Answer 1

Answer: C) -348J

Step-by-step explanation:

42.741 - 390.76 (the kinetic energy of the two speeds) is equal to -348.

Answer 2

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

Step-by-step explanation:

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

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

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

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

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

The energy change was approximately:

`390.76\; {\rm J} - 42.741\; {\rm J} \approx 348\; {\rm J}`.

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