Question

two forces are acting on the body of mass 10 kg in opposite direction . one force is of 100 N if frictional force is 10 N and body is moving in the direction of application of first force with acceleration of 2 m/s2 . find another force acting on body​

Answer 1

$\Large\underline{\underline{\red{\sf \purple{\maltese}\:\: Given :- }}}$

$\sf\implies Two\ opposing\ forces \ are \ acting \ on \ a \ body ( 10kg ). \\\implies \textsf{ One force is 100N and frictional force is 10N . }\\\implies \textsf{ Acceleration of the body is \sf 2m/s^2 .}$

$\Large\underline{\underline{\red{\sf \purple{\maltese}\:\: To \ Find :- }}}$

$\implies\textsf{ The other force acting on the body. }$

$\Large\underline{\underline{\red{\sf \purple{\maltese}\:\: Answer :- }}}$

Let us assume the second force to be F . Since the body is moving in the direction of application of first force , the second force and the frictional force will be opposing in nature. Hence the net opposing force will be ( F + 10 ) N.

$\underline{\underline{\purple{\sf Diagram :- }}}$

$\setlength{\unitlength}{1 cm}\begin{picture}(12,12)\thicklines\put( 0,0){\line(1,0){6}} \put(3,0){\line(0,1){1}}\put(3,1){\line(1,0){1}}\put(4,1){\line(0,-1){1}}\put(2,0.5){\vector(1,0){1}}\put(5,0.5){\vector(-1,0){1}}\put(2,0.8){ \bf 100N } \put(4.5,0.8){ \bf F } \put(3,0.5){\bf 10kg} \put(0.2,0){\line(-1,-1){0.36}}\put(0.4,0){\line(-1,-1){0.36}}\put(0.6,0){\line(-1,-1){0.36}}\put(0.8,0){\line(-1,-1){0.36}}\put(1,0){\line(-1,-1){0.36}}\put(1.2,0){\line(-1,-1){0.36}}\put(1.4,0){\line(-1,-1){0.36}}\put(1.6,0){\line(-1,-1){0.36}}\put(1.8,0){\line(-1,-1){0.36}}\put(3,0){\line(-1,-1){0.36}}\put(2.2,0){\line(-1,-1){0.36}}\put(3.2,0){\line(-1,-1){0.36}}\put(2.4,0){\line(-1,-1){0.36}}\put(2.6,0){\line(-1,-1){0.36}}\put(2.8,0){\line(-1,-1){0.36}}\put(3.4,0){\line(-1,-1){0.36}}\put(2,0){\line(-1,-1){0.36}}\put(3.6,0){\line(-1,-1){0.36}}\put(3.8,0){\line(-1,-1){0.36}}\put(4,0){\line(-1,-1){0.36}}\put(4.2,0){\line(-1,-1){0.36}}\put(4.4,0){\line(-1,-1){0.36}}\put(4.6,0){\line(-1,-1){0.36}}\put(4.8,0){\line(-1,-1){0.36}}\put(5,0){\line(-1,-1){0.36}}\put(5.2,0){\line(-1,-1){0.36}}\put(5.4,0){\line(-1,-1){0.36}}\put(5.6,0){\line(-1,-1){0.36}}\put(5.8,0){\line(-1,-1){0.36}}\put(6,0){\line(-1,-1){0.36}}\put(5,-0.6){\vector(-1,0){1 }}\put(5,-0.7){\bf 10N }\end{picture}$

$\underline{\underline{\orange{\boldsymbol{According \ to \ the \ Question :- }}}}$

$\sf:\implies \pink{ Force = mass \times Acceleration} \\\\\sf:\implies F_{net} = 10kg \times 2m/s^2 \\\\\sf:\implies 100N - ( F + 10 ) N = 20 kg - m/s^2 \\\\\sf:\implies 100 N- F - 10 N= 20 N\\\\\sf:\implies 90N - F = 20 N \\\\\sf:\implies F = 90N - 20 N \\\\:\implies \boxed{\pink{\mathfrak{ Force = 70 Newtons }}}$

$\underline{\blue{\sf \therefore Hence \ the \ second \ force \ is \ \textsf{\textbf{70\ Newtons }}. }}$