Question

An object moving horizontally with a kinetic energy of 800 J experiences a constant opposing force of 100 N while moving from one point X to another point Y, where XY is 2 m. What is the energy of the object at Y?

Answer 1

$$\begin{gathered} Letmassofthebodybe\text{'}m\text{'}.Then \\ \frac{1}{2}m{v_1}^2=800 \\ {v_1}^2=\frac{1600}{m}---\left(1\right) \\ and, \\ ma=100 \\ a=-\frac{100}{m}(\sin ceitisoppo\sin gthemotion)----\left(2\right) \\ fromthirdequationofmotion \\ {v_2}^2={v_1}^2+2as \\ {v_2}^2=\frac{1600}{m}-2\times \frac{100}{m}\times 2 \\ {v_2}^2=\frac{1600}{m}-\frac{400}{m} \\ {v_2}^2=\frac{1200}{m} \\ or, \\ m{v_2}^2=1200 \\ multiplyingbothsidesby\frac{1}{2} \\ \frac{1}{2}m{v_2}^2=\frac{1}{2}\times 1200 \\ \frac{1}{2}m{v_2}^2=600JwhichisthekineticenergyofthebodyatpointY \end{gathered}$$