Question

a 12 kg bag of groceries is tossed onto a table at 3.0 m/s and slides to a stop in 3.0 s. part a modify the equation fδt = δ (mv) to find the force of friction. express your answer to two significant figures and include the appropriate units. f = value units

Explanation:

Step1: Identify the initial and final momentum

The initial momentum $p_i = mv_i$, where $m = 12\ kg$ and $v_i=3.8\ m/s$. The final momentum $p_f = 0$ (since it comes to a stop).
$p_i=12\times3.8 = 45.6\ kg\cdot m/s$ and $\Delta p=p_f - p_i=0 - 45.6=- 45.6\ kg\cdot m/s$

Step2: Use the impulse - momentum theorem

The impulse - momentum theorem is $F\Delta t=\Delta p$. We know $\Delta t = 3.0\ s$. Rearranging for $F$, we get $F=\frac{\Delta p}{\Delta t}$.
$F=\frac{- 45.6}{3.0}=-15.2\ N$. The magnitude of the force is considered, so $F = 15\ N$ (to two significant figures)

Answer:

15 N