A block of mass 2.0 kg is pushed down an inclined plane of inclination `37^0` with a force of 20N acting parallel to the incline. It is found that the block moves on the incline with an acceleration of `10 m/s^2`. If the block started from rest, find the work done by the applied force in the first second .

Answer

A, B, C

Explanation

`m=2kg, theta=37^0, F=20N, a=10m/sec^2`
at=1sec
So, `s=ut+1/2at^2=5m`

Work dne by the applied force
`W=Fscostheta^0=20xx5=100J`
b. `BC(h)=5sin37^0=3m`
so, work done by weight ltbr. `W=mgh`
2xx10xx3=60J`
So, frictioN/Al force,
`f=mgsintheta`
Work done by the friction forces
`W=fs costheta0^0=(mg sinthega)s`
`=20xx0.60xx5=-60J`