Answered

IDNLearn.com: Your one-stop destination for finding reliable answers. Discover comprehensive answers from knowledgeable members of our community, covering a wide range of topics to meet all your informational needs.

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, the box (initially at rest) is traveling at a speed of 6 m/s. How much work (in Joules) did friction do in this process

Sagot :

Answer: 321 J

Explanation:

Given

Mass of the box [tex]m=3\ kg[/tex]

Force applied is [tex]F=25\ N[/tex]

Displacement of the box is [tex]s=15\ m[/tex]

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

acceleration associated with it is [tex]a=\dfrac{F}{m}[/tex]

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

Work done by force is [tex]W=F\cdot s[/tex]

[tex]W=25\times 15\\W=375\ J[/tex]

change in kinetic energy is [tex]\Delta K[/tex]

[tex]\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[/tex]

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

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

Therefore, the magnitude of work done by friction is [tex]321\ J[/tex]

Thank you for using this platform to share and learn. Don't hesitate to keep asking and answering. We value every contribution you make. IDNLearn.com has the solutions to your questions. Thanks for stopping by, and come back for more insightful information.