Question

Hola mates!!


Give answer ✌➡

A body,when acted upon by force of 10kgf,gets displaced by 0.5m. Calculate the work done by the force ,when the displacement is ➡__➡__➡__➡__➡__➡__➡__

(i)in the direction of force.

(ii)at an angle of 60° with the force and

(iii)normal to the force.


✊Answerz✊

(i)50J

(ii)25J

(iii) 0


✌✌


#Thankx#

Answer 1

Work done = force ×displacement ×cosa
1) w= 10×10×0.5
= 50joule
2)w= 10×10×0.5×cos60
=25 joules
3) cos 90 =0 so work done is zero

Answer 2

Hey mate!!



✌Your answerz✌


Force➡10kgf =10*10 =100N

Displacement ➡0.5m

(i) Work done=Force*displacement

=100*0.5

=50J

(ii)Work= F*S*CosO

=100*0.5*60°

=50*0.5(because 0.5 is a value of Cos60)

=25J

(iii) W= F*S*CosO

=100*0.5*0

=0



Hence all right answers ✌



☺Thank you☺



❤ Follow me for more answers ❤