The correct option is C 2250 J
Here for finding the workdone we can take the component of force along the displacement or component of displacement along the direction of force.
Method (I)
Taking component of force(mgsinθ) along the displacement(s)
Displacement along incline, |→s|=12cos37∘, as cos37o=45
|→s|=12×54=15 m
Force due to gravity =mg=25×10=250 N,
Component of force along the incline is mgsinθ=250×35=150 N
Hence, workdone W=F×S=150×15=2250 J, as the force is along the displacement (cos0o=0)
Method (II)
Taking the component of displacement(h) along force (mg)
By the figure, tan37o=h12
⇒h=12×34=9 m
Hence work done, W=F×h=mg×h=250×9=2250 J
Option C is the correct answer