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.

Dear student, 

From Work energy theorem,
Total Work done by block in coming from A to B = Change in Kinetic energy .
i.e, 
WFriction+Wgravity = K.EWFriction +MgH     = Kfinal- Kinitial WFriction   = 12 m V2final- 12 m V2initial - MgH      (i)Here H = 6 m , g =10m/s2(assumed) , Vfinal=8m/s , Vinitial=1.2m/s As F = 300N , g =10m/s2M =Fg =30010 =30Kg.Putting all gievn values  WFriction   =12 m V2final-  V2initial - MgH                   =12 ×30  82-  1.22 - 30 × 10 × 6                 =- 861.6 N
Negative sign shows that Work done by friction is opposing the motion.

Regards

  • 12
change in kinetic energy is equal to sum of potential enegy and frictional energy loss
  • -2
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