The 300 N crate slides down the inclined curved path from A to B in the vertical plane as shown - Physics - Reproduction in Organisms

Question

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

Adarsh Dubey · Accepted Answer

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

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.