Question

question 12
a net force of 25 n is applied for 5.7 s to a 12 - kg box initially at rest. what is the speed of the box at the end of the 5.7 - s interval? (hint: find acceleration first, then use equations of kinematics from chapter 2)
12 m/s
3.0 m/s
30 m/s
7.5 m/s
1.8 m/s

Explanation:

Step1: Calculate acceleration

According to Newton's second - law $F = ma$, so $a=\frac{F}{m}$. Given $F = 25\ N$ and $m = 12\ kg$, then $a=\frac{25}{12}\ m/s^{2}\approx2.083\ m/s^{2}$.

Step2: Use kinematic equation

The kinematic equation $v = v_0+at$. Since the box is initially at rest, $v_0 = 0\ m/s$, $a=\frac{25}{12}\ m/s^{2}$, and $t = 5.7\ s$. Then $v=0+\frac{25}{12}\times5.7=\frac{25\times5.7}{12}=\frac{142.5}{12}=11.875\ m/s\approx12\ m/s$.

Answer:

12 m/s