Question 7.13: A luge and its rider, with a total mass of 85 kg, emerge fro......
A luge and its rider, with a total mass of 85 kg, emerge from a downhill track onto a horizontal straight track with an initial speed of 37 m/s. If a force slows them to a stop at a constant rate of 2.0 m/s², (a) what magnitude F is required for the force, (b) what distance d do they travel while slowing, and (c) what work W is done on them by the force? What are (d) F, (e) d, and (f) W if they, instead, slow at 4.0 m/s²?
We choose +x as the direction of motion (so \vec{a} and \vec{F} are negative-valued).
(a) Newton’s second law readily yields \vec{F}=(85\, kg )\left(-2.0\, m / s ^2\right) so that
F=|\vec{F}|=1.7 \times 10^2 \,N.
(b) From Eq. 2-16 (with v = 0) we have
\nu^2=\nu_0^2+2 a\left(x-x_0\right), (2-16)
0=\nu_0^2+2 a \Delta x \Rightarrow \Delta x=-\frac{(37 \,m / s )^2}{2\left(-2.0 \,m / s ^2\right)}=3.4 \times 10^2\, m.
Alternatively, this can be worked using the work-energy theorem.
(c) Since \vec{F} is opposite to the direction of motion (so the angle \phi between \vec{F} and \vec{d}=\Delta x is 180°) then Eq. 7-7 gives the work done as W=-F \Delta x=-5.8 \times 10^4 \,J .
W=F d \cos \phi (work done by a constant force). (7-7)
(d) In this case, Newton’s second law yields \vec{F}=(85\,kg )\left(-4.0\, m / s ^2\right) so that F=|\vec{F}|=3.4 \times 10^2\, N \text{.}
(e) From Eq. 2-16, we now have
\Delta x=-\frac{(37 \,m / s )^2}{2\left(-4.0 \,m / s ^2\right)}=1.7 \times 10^2 \,m.
(f) The force \vec{F} is again opposite to the direction of motion (so the angle \phi is again 180°) so that Eq. 7-7 leads to W = -FΔx = -5.8×10^4 J. The fact that this agrees with the result of part (c) provides insight into the concept of work.