IDNLearn.com is designed to help you find accurate answers with ease. Get accurate answers to your questions from our community of experts who are always ready to provide timely and relevant solutions.
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]