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]

1 Answer
Sep 1, 2017

#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.