# The 300 N crate slides down the inclined curved path from A to B in the vertical plane as shown in the figure. If the crate has a velocity of 1.2 m/s down the inclined at point A and 8 m/s at B. Find the work done against friction during the motion.

From Work energy theorem,

Total Work done by block in coming from A to B = Change in Kinetic energy .

i.e,

${W}_{Friction}+{W}_{gravity}=\u2206K.E\phantom{\rule{0ex}{0ex}}{W}_{Friction}+MgH={K}_{final}-{K}_{initial}\phantom{\rule{0ex}{0ex}}{W}_{Friction}=\left[\frac{1}{2}m{{V}^{2}}_{final}-\frac{1}{2}m{{V}^{2}}_{initial}\right]-MgH\left(i\right)\phantom{\rule{0ex}{0ex}}HereH=6m,g=10m/{s}^{2}\left(assumed\right),{V}_{final}=8m/s,{V}_{initial}=1.2m/s\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}AsF=300N,g=10m/{s}^{2}\phantom{\rule{0ex}{0ex}}M=\frac{F}{g}=\frac{300}{10}=30Kg.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}Puttingallgievnvalues\phantom{\rule{0ex}{0ex}}{W}_{Friction}=\frac{1}{2}m\left[{{V}^{2}}_{final}-{{V}^{2}}_{initial}\right]-MgH\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\times 30\left[{8}^{2}-1.{2}^{2}\right]-30\times 10\times 6\phantom{\rule{0ex}{0ex}}=-861.6N\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

Negative sign shows that Work done by friction is opposing the motion.

Regards

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