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]
A cyclist and her bicycle have a total mass of
1 Answer
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
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
#"F"_"uphill" = m * a#
This will get you
#m * a = P/v_"uphill" - R - "mgsin(theta)" " " "color(blue)((1))#
#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
Once again, you have
#m * a = P/v_"downhill" - R + mgsin(theta)" " " "color(blue)((2))#
You now have two equations with two unknowns,
Combine equations
#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
#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
Using equation
#"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.