An object of mass 500 ~g , initially at rest acted upon by a variable force whose X component varies with X in the manner shown. The velocities of the object a point X=8 ~m and X=12 ~m , would be the respective values of (nearly)

Step-by-step solution

The area under the force displacement curve give the amount of work done. From work-energy theorem, aligned & W = KE & At x =8 ~m , & W = Area ABDO + Area CEFD & =20 5+10 3=130 ~J & Using Eq. (i) & 130= 1 2 mv ^2= 1 2 500 1000 v ^2 & v =2 130 =22.8 ~ms ⁻¹ 23 ~ms ⁻¹ & At x =12 ~m & ~W = Area ABDO + Area CEFD + Area FGHIJ & + Area KLMJ & W =20 5+10 3+(-20 2)+ ( 1 2 -5 2 ) & +10 2 [ Area FGHIJ = Area & FGIJ + Area GHI ] & =100+30-40-5+20=105 ~J & Using Eq. (i) & 105= 1 2 1 2 v^2 & v =2 105 20.6 ~ms ⁻¹ & aligned