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AlexFokin [52]
3 years ago
5

Suppose you are pushing a 3 kg box with a force of 25 N (directed parallel to the ground) over a distance of 15 m. Afterward, th

e box (initially at rest) is traveling at a speed of 6 m/s. How much work (in Joules) did friction do in this process
Physics
1 answer:
alex41 [277]3 years ago
3 0

Answer: 321 J

Explanation:

Given

Mass of the box m=3\ kg

Force applied is F=25\ N

Displacement of the box is s=15\ m

Velocity acquired by the box is v=6\ m/s

acceleration associated with it is a=\dfrac{F}{m}

\Rightarrow a=\dfrac{25}{3}\ m/s^2

Work done by force is W=F\cdot s

W=25\times 15\\W=375\ J

change in kinetic energy is \Delta K

\Rightarrow \Delta K=\dfrac{1}{2}m(v^2-0)\\\\\Rightarrow \Delta K=\dfrac{1}{2}\times 3\times 6^2\\\\\Rightarrow \Delta K=\dfrac{1}{2}\times 3\times 36\\\\\Rightarrow \Delta K=54\ J

According to work-energy theorem, work done by all the forces is equal to the change in the kinetic energy

\Rightarrow W+W_f=\Delta K\quad [W_f=\text{Work done by friction}]\\\\\Rightarrow 375+W_f=54\\\Rightarrow W_f=-321\ J

Therefore, the magnitude of work done by friction is 321\ J

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