Question

A block of weight 100 N is slowly slid up on a smooth incline of inclination 37° by a person. Calculate the work done by the person in moving the block through a distance of 2.0 m, if the driving force is (a) parallel to the incline and (b) in the horizontal direction.
Concept of Physics - 1 , HC VERMA , Chapter " Work and Energy"

Answer 1

HEY!!

_______________________________

✴Weight, mg=100 N

✴θ= 37° and s= 2 m

✴Force, F=mg sin 37°

✴100 ×35= 60 N

✴W=FS cos θ

✴60×2×cos 0° =120 J

✴In ΔABC, AB=2 m

✴AC=h

✴s×sin 37°=2.0×sin 37°

✴1.2 m

▶Work done when the force is in horizontal direction

▶W'=mgh

✔✔100×1.2=120 J

Answer 2

(a) Parallel to the incline :-

Have a Look to the attachment diagram:-
 here the block is slowly sliding up,
so driving force parallel to incline = resisting force of component of weight
given :-
w = 100 N
θ = 37°
distance s = 2.0 m

Then driving force F = 100 × sin37° N
∴ Work done = F × d
Work done = 100 × sin37° × 2.0 m
=120 J

Hence the work done in the parallel incline is 120 J.


(b) Force in the horizontal direction.

When the driving force occurs in the horizontal direction , along incline its component is equal to the force F which is 100 × sin37° N
So in this case also,
The work done = 120 J. 


Hence the work done in the horizontal direction is 120 J.


Hope it Helps