Answer to Question #178390 in Physics for Debbie Daniel

Question

A loaded sled weighing 1,250 lb is given a speed of 25.0 mi/h while 

moving a distance of 140 ft from rest on a horizontal ice surface. If the coefficient of friction 

is 0.105, what constant force, applied horizontally, would be necessary to produce this 

motion?

Answer 1

The net force, acting on the sled is $F - F_f$, where $F$ is the applied force, and $F_f = \mu m g$ is the force of friction ($\mu$ is the coefficient of friction).

Hence, according to 2^nd Newton's law, $m a = F - F_f = F - \mu m g$, from where $F = m (a + \mu g)$.

The distance for an accelerated motion can be calculated as $s = \frac{v^2-v_0^2}{2 a}$(here $v_0 = 0$ is the initial speed and $v = 25 \frac{mi}{h}$ is the speed after covering the distance $s = 140 ft$). From the last equation, acceleration is $a = \frac{v^2}{2 s}$.

Substituting the acceleration into the expression for the force, obtain:

$F = m \left(\frac{v^2}{2 s} + \mu g\right) \approx 1413 N$.