An object travels $7.5 \mathrm{~m} / \mathrm{s}$ toward the west. Under the influence of a constant net force of $5.2 \mathrm{kN}$, after $4.3 \mathrm{~s}$, it is travelling $2.5 \mathrm{~m} / \mathrm{s}$ toward the east. What is its mass?
A) $2200 \mathrm{~kg}$
B) $4500 \mathrm{~kg}$
C) $12000 \mathrm{~kg}$
D) $6000 \mathrm{~kg}$

Question

Quizplus · Accepted Answer

The change in velocity is $7.5 \mathrm{~m/s} + 2.5 \mathrm{~m/s} = 10 \mathrm{~m/s}$ (since it changes direction to the east). Using the formula $F = ma$ and $a = \Delta v / \Delta t$, where $\Delta v = 10 \mathrm{~m/s}$ and $\Delta t = 4.3 \mathrm{~s}$, we find $a = 10 / 4.3 \approx 2.33 \mathrm{~m/s^2}$. Given $F = 5200 \mathrm{~N}$ (since $1 \mathrm{kN} = 1000 \mathrm{N}$), mass $m = F / a = 5200 / 2.33 \approx 2232 \mathrm{~kg}$, which is closest to $2200 \mathrm{~kg}$.

An Object Travels 7.5 m/s7.5 \mathrm{~m} / \mathrm{s}7.5 m/s Toward the West 5.2kN5.2 \mathrm{kN}5.2kN

An Object Travels $7.5 \mathrm{~m} / \mathrm{s}$ Toward the West $5.2 \mathrm{kN}$