Question

a man pulls a 48.0 kg box horizontally from rest while exerting a constant horizontal force, displacing the box 2.60 meters in 2.00 seconds. find the force the man exerts on the box. (ignore friction.)

Explanation:

Step1: Find the acceleration

Use the kinematic - equation $x = v_0t+\frac{1}{2}at^{2}$. Since the box starts from rest, $v_0 = 0$. The equation simplifies to $x=\frac{1}{2}at^{2}$. We can solve for acceleration $a$:
\[a=\frac{2x}{t^{2}}\]
Given $x = 2.60\ m$ and $t = 2.00\ s$, then $a=\frac{2\times2.60}{2.00^{2}}=\frac{5.2}{4}=1.3\ m/s^{2}$.

Step2: Calculate the force

According to Newton's second - law $F = ma$. Given $m = 48.0\ kg$ and $a = 1.3\ m/s^{2}$, then $F=48.0\times1.3 = 62.4\ N$.

Answer:

$62.4$