Mailosi Namacha

2022-06-30

a horizontal force is applied to a 4kg box, which is initially at rest on a rough horizontal table.The coefficient of kinetic friction between the box and the table is 0.35 find the speed of the box after it has been pushed 3m

To solve the problem, we can use the principles of Newtonian mechanics and the concept of work and energy. Let's break it down step by step:
Given:
Mass of the box, $m=4\phantom{\rule{0.167em}{0ex}}\text{kg}$
Coefficient of kinetic friction, ${\mu }_{k}=0.35$
Distance pushed, $d=3\phantom{\rule{0.167em}{0ex}}\text{m}$
We can start by calculating the work done on the box by the applied force. The work done is equal to the force applied multiplied by the distance moved. The force applied can be calculated using the equation $F={\mu }_{k}·N$, where $N$ is the normal force acting on the box.
The normal force $N$ is equal to the weight of the box, which is given by $mg$, where $g$ is the acceleration due to gravity. Therefore, $N=mg$.
Using the given mass $m=4\phantom{\rule{0.167em}{0ex}}\text{kg}$, and assuming $g=9.8\phantom{\rule{0.167em}{0ex}}{\text{m/s}}^{2}$, we can calculate the normal force $N$ as:
$N=mg=\left(4\phantom{\rule{0.167em}{0ex}}\text{kg}\right)\left(9.8\phantom{\rule{0.167em}{0ex}}{\text{m/s}}^{2}\right)=39.2\phantom{\rule{0.167em}{0ex}}\text{N}$
Now, we can calculate the force of kinetic friction $F$:
$F={\mu }_{k}·N=\left(0.35\right)\left(39.2\phantom{\rule{0.167em}{0ex}}\text{N}\right)=13.72\phantom{\rule{0.167em}{0ex}}\text{N}$
The work done on the box is given by the equation:
$W=F·d$
Substituting the values, we have:
$W=\left(13.72\phantom{\rule{0.167em}{0ex}}\text{N}\right)\left(3\phantom{\rule{0.167em}{0ex}}\text{m}\right)=41.16\phantom{\rule{0.167em}{0ex}}\text{J}$
The work done on the box is equal to the change in kinetic energy of the box. Therefore, we can express this as:
$W=\Delta KE$
The change in kinetic energy can be calculated using the equation:
$\Delta KE=\frac{1}{2}m{v}_{f}^{2}-\frac{1}{2}m{v}_{i}^{2}$
Where ${v}_{f}$ is the final velocity (speed) of the box, and ${v}_{i}$ is the initial velocity (speed), which is zero since the box starts from rest.
Since ${v}_{i}=0$, the equation simplifies to:
$\Delta KE=\frac{1}{2}m{v}_{f}^{2}$
Substituting the known values, we have:
$41.16\phantom{\rule{0.167em}{0ex}}\text{J}=\frac{1}{2}\left(4\phantom{\rule{0.167em}{0ex}}\text{kg}\right){v}_{f}^{2}$
Simplifying the equation, we find:
${v}_{f}^{2}=\frac{41.16\phantom{\rule{0.167em}{0ex}}\text{J}}{2\phantom{\rule{0.167em}{0ex}}\text{kg}}$
${v}_{f}^{2}=20.58\phantom{\rule{0.167em}{0ex}}{\text{m}}^{2}/{\text{s}}^{2}$
Finally, to find the speed of the box, we take the square root of both sides:
${v}_{f}=\sqrt{20.58\phantom{\rule{0.167em}{0ex}}{\text{m}}^{2}/{\text{s}}^{2}}\approx 4.53\phantom{\rule{0.167em}{0ex}}\text{m/s}$
Therefore, the speed of the box after it has been pushed 3m is approximately $4.53\phantom{\rule{0.167em}{0ex}}\text{m/s}$.

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