Question

A force of 15 N is required to pull up a body of
mass 2 kg through a distance 5 m along an
inclined plane making an angle of 30° with the
horizontal as shown in Fig. 2.11. Calculate :
(i) the work done by the force in pulling the
body,
(ii) the force due to gravity on the body,
(iii) the work done against the force due to gravity.
Take : g = 9.8 ms?
(iv) Account for the difference in answers of
part (1) and part (iii).


WRONG ANSWERS WILL BE REPORTED. mind it !! only who know will answer​

Answer 1

1) Work done by the force in pulling the body up

W = Force x displacement in the direction of

force = 15 N x 5 m = 75 J

(ii) Force due to gravity on the body

Famg = 2 x 9.8 = 19-6 N

(ii) Work done against the force due to

gravity W' = Force due to gravity vertical

height moved

= mg x BC

But in right angled A ACB, sin 30º =

- BC = AB sin 30°

Hence W' = mg x AB sin 30°

= 196 x 5 x

( sin 30° = 17)

(iv) We note that W > W'. The difference in work W

and W' is 75 J - 49 J = 26 J. Actually 26 J is

the work done against the force of friction

between the body and the inclined plane.

Answer 2

Answer:

given:

θ = 30°

F = 15 N

m = 2 kg

d = 5 m

( i ) work done =>  W = F d cosθ

     W = 15 * 5 * cos 30°

     W = 64.95  ≅  65 J

( ii )  $F_g$ = m*g

      $F_g$ = 2*9.8

     $F_g$  = 19.6 N

( iii )   work done against the gravity = mgh

                                                            = 2*9.8*2.5    ( h = d sinθ )

                                                            = 49 J

( iv ) difference in answers of ( i ) and ( iii ) is 16 J

in the work done ( 65 J ) by the force 49 J is used to nullify the force due to gravity on the body.

And then 16 J is used to take the body to a height of 2.5 m from the ground ( 5 m on the inclined plane ).

Hope it helps !