Question

Finding Power and Resistance?

A cyclist and her bicycle have a total mass of `84kg`. She works at a constant rate of `P` `W` while moving on a straight road which is inclined to the horizontal at an angle `theta`, where `sintheta = 0.1`. When moving uphill, the cyclist's acceleration is `1.25m`/`s^2` at an instant when her speed is `3m`/`s^1`. When moving downhill, the cyclist's acceleration is `1.25m`/`s^2` at an instant when her speed is `10m`/`s^1`. The resistance to the cyclist's motion, whether the cyclist is moving uphill or downhill, is `R` `N`. Find the values of `P` and `R`. [8]

Answer 1

`P = "720 W"`

`R = "51 N"`

Explanation:

Your starting point here will be to find a relationship between the power exerted by the cyclist and her velocity at the two instances.

You know that

`"power" = "work"/"time"`

and

`"work" = "force" * "displacement" * cos(alpha)`

Now, it's very important to realize that you have

`alpha = 0^@`

and

`cos(alpha) = 1`

This is the case even though the cyclist is moving on a surface that is inclined to the horizontal at an angle `theta` for which `sin(theta) = 0.1`, the angle between the net force that's acting on the cyclist and the direction of the movement is `0^@`.

In other words, the net force is parallel to the road.

You can thus say that you have

`"work" = "force" * "displacement"`

Now, you also know that

`"displacement" = "velocity" * "time"`

Plug this into the equation you have for power to get

`"power" = ("force" * "velocity" * color(red)(cancel(color(black)("time"))))/color(red)(cancel(color(black)("time")))`

This means that you have

`"power" = "velocity" * "force"`

Consequently, you can say that

`"force" = "power"/"velocity"`

Next, focus on finding the net force that is acting on the cyclist on her way uphill and on her way downhill*.

• `ul("moving uphill")`

If we take the direction of movement to be the positive direction, you can say that

`"F"_ "uphill" = P/v_"uphill" - R - "mgsin(theta)`

Here

• `P` is the power exerted by the cyclist
• `v_"uphill"` is her velocity when her acceleration is `"1.25 m s"^(-2)`
• `R` is the resistance
• `m` is the mass of the cyclist

Since you know that the cyclist has an acceleration of `"1.25 m s"^(-2)` while going uphill, you can say that

`"F"_"uphill" = m * a`

This will get you

`m * a = P/v_"uphill" - R - "mgsin(theta)" " " "color(blue)((1))`

`color(white)(a)`

• `ul("moving downhill")`

Once again, if we take the direction of movement to be the positive direction, you can say that

`F_"downhill" = P/v_"downhill" - R + mgsin(theta)`

Notice that `R` is opposing the motion both uphill and downhill, but this time, the tangential component of the weight is aiding the movement, hence why it carries a `+` sign.

Once again, you have `a = "1.25 m s"^(-2)`, so

`m * a = P/v_"downhill" - R + mgsin(theta)" " " "color(blue)((2))`

You now have two equations with two unknowns, `P` and `R`.

Combine equations `color(blue)((1))` and `color(blue)((2))` to get

`P/v_"uphill" - color(red)(cancel(color(black)(R))) - mgsin(theta) = P/v_"downhill" - color(red)(cancel(color(black)(R))) + mgsin(theta)`

`P * (1/v_"uphill" - 1/v_"downhill") = 2mgsin(theta)`

This is equivalent to

`P = (2mgsin(theta))/(1/v_"uphill" - 1/v_"downhill")`

Plug in your values to find--to make the calculations easier, I'll take `g = "10 m s"^(-2)`. Since `sin(theta) = 0.1`, this will get `g * sin(theta) = 1`

`P = (2 * "84 kg" * "10 m s"^(-2) * 0.1)/(1/("3 m s"^(-1)) - 1/("10 m s"^(-1)))`

`P = color(darkgreen)(ul(color(black)("720 kg m"^2 "s"^(-3) = "720 W")))`

I'll leave the answer rounded to two sig figs.

Now that you know the value of `P`, pick one of the two equations and solve for `R`.

Using equation `color(blue)((2))`, you have

`"84 kg" * "1.25 m s"^(-2) = ("720 kg m"^color(red)(cancel(color(black)(2))) "s"^color(red)(cancel(color(black)(-3))))/(10 color(red)(cancel(color(black)("m"))) "s"^color(red)(cancel(color(black)(-1)))) - R + "84 kg" * "10 m s"^(-2) * 0.1`

This will get you

`R = (72 + 84 * 10 * 0.1 - 84 * 1.25) color(white)(.)"kg m s"^(-2)`

`R = color(darkgreen)(ul(color(black)("51 kg m s"^(-2) = "51 N")))`

Once again, I'll leave the answer rounded to two sig figs.