If we assume the acceleration that the sled undergoes is constant throughout its motion, then we have the average velocity [tex]\bar v[/tex] of the sled to be
[tex]\bar v=\dfrac{v_i+v_f}2=\dfrac{\Delta x}{23\,\mathrm s}[/tex]
where [tex]\Delta x[/tex] is the total displacement of the sled, and [tex]v_i[/tex] and [tex]v_f[/tex] are the sled's initial and final velocities, respectively. The sled eventually stops, so we take [tex]v_f=0[/tex] and solve for [tex]v_i[/tex]:
[tex]\dfrac{v_i}2=\dfrac{15\,\mathrm m}{23\,\mathrm s}\implies v_i=1.3\,\dfrac{\mathrm m}{\mathrm s}[/tex]
Now, take the sled's starting position to be the origin. The sled moves in one direction, which we take to be the positive direction. Then because it's slowing down, we expect its acceleration to be in the negative direction (and hence with negative sign). In particular, the sled's position [tex]x[/tex] at time [tex]t[/tex] is
[tex]x=x_i+v_it+\dfrac12at^2[/tex]
We have [tex]\Delta x=x-x_i=15\,\mathrm m[/tex], [tex]v_i=1.3\,\frac{\mathrm m}{\mathrm s}[/tex], and [tex]t=23\,\mathrm s[/tex], so we can solve for acceleration [tex]a[/tex]:
[tex]15\,\mathrm m=\left(1.3\,\dfrac{\mathrm m}{\mathrm s}\right)(23\,\mathrm s)+\dfrac12a(23\,\mathrm s)^2[/tex]
[tex]\implies a=-0.056\,\dfrac{\mathrm m}{\mathrm s^2}[/tex]
With a mass of [tex]m=52.5\,\mathrm{kg}[/tex], we find that the stopping force is
[tex]F=ma=-2.9\,\mathrm N[/tex]
which means the stopping force has magnitude [tex]2.4\,\mathrm N[/tex] in the negative direction (opposite the direction of the sled's initial velocity).
