Question

solution
find the mass of the system from the definition of weight, w = mg.
m=\frac{w}{g}=\frac{3.00\times10^{2} n}{9.80 m/s^{2}} = 30.6 kg
find the acceleration of the system from the second law.
a_{x}=\frac{f_{x}}{m}=\frac{20.0 n}{30.6 kg}=0.654 m/s^{2}
use kinematics to find the distance moved in 2.00 s, with (v_{0}=0).
delta x=\frac{1}{2}a_{x}t^{2}=\frac{1}{2}(0.654 m/s^{2})(2.00 s)^{2}=1.31 m
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remarks note that the constant applied force of 20.0 n is assumed to act on the system at all times during its motion. if the force were removed at some instant, the system would continue to move with constant velocity and hence zero acceleration. the rollers have an effect that was neglected here.
question if the weight of the crate were doubled, the acceleration would be multiplied by
and the displacement attained in 2.0 s would be multiplied by
practice it
use the worked example above to help you solve this problem. the combined weight of the crate and dolly as shown in the figure is (2.90\times10^{2} n). if the man pulls on the rope with a constant force of 19.5 n, what is the acceleration of the system (crate plus dolly), and how far will it move in 2.00 s? assume that the system starts from rest and that there are no friction forces opposing the motion.
acceleration (m/s^{2})
displacement (m)
exercise
hints: getting started | im stuck!
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.)
n

Explanation:

Step1: Calculate mass from weight

First, find the mass $m$ of the system using $m=\frac{w}{g}$, where $w = 2.90\times10^{2}\ N$ and $g = 9.80\ m/s^{2}$. So, $m=\frac{2.90\times 10^{2}\ N}{9.80\ m/s^{2}}\approx29.6\ kg$.

Step2: Calculate acceleration

Then, use Newton's second - law $a=\frac{F}{m}$ to find the acceleration. Given $F = 19.5\ N$ and $m\approx29.6\ kg$, so $a=\frac{19.5\ N}{29.6\ kg}\approx0.659\ m/s^{2}$.

Step3: Calculate displacement

Finally, use the kinematic equation $\Delta x=v_{0}t+\frac{1}{2}at^{2}$. Since $v_{0} = 0\ m/s$, $a\approx0.659\ m/s^{2}$ and $t = 2.00\ s$, we have $\Delta x=\frac{1}{2}(0.659\ m/s^{2})(2.00\ s)^{2}=1.32\ m$.

Answer:

acceleration: $0.659\ m/s^{2}$
displacement: $1.32\ m$