A block of mass 5 kg is pulled along a horizontal. The friction coefficient between the block and the surface is `0.5`. If the block travels at constant velocity, find the work done by this applied force during a displacement 4 m of the block.

Explanation

`uarr:N_0+Fsin37^@=50`
`N_0=50-(3F)/(5)`
`f=muN_0=(1)/(2)(50-(3F)/(5))=25-(3F)/(10)`
`rarr`: Since the block is moving with constant velocity `Fcos37^@=f`
`(4F)/(5)=25-(3F)/(10)implies(11F)/(10)=25impliesF=(250)/(11)N`
Work done by the applied force is
`Fcos37^@d=(250)/(11)xx(4)/(5)xx4=(800)/(11)J`