34. A body of mass 2kg has an initial speed 5 msforce acts on it for some time in the direction ofmotion. The force time graph is shown in figure,The final speed of the body is. a. 8.5m/s. b. 11m/s. c. 14.31m/s. d.4.31m/s
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Figure is not given in your question
Impulse=Change in momentum =m(v2−v1)=m(v2-v1) …(i)
Again impulse=Area between the graph and time axis
=12×2×4+2×4+12(4+2.5)×0.5+2×2.5=12×2×4+2×4+12(4+2.5)×0.5+2×2.5
=4+8+1.625+5=18.625=4+8+1.625+5=18.625
From (i) and (ii), m(v2−v1)=18.625m(v2-v1)=18.625
⇒v2=18.625m=v1=18.6252+5=14.25m/s⇒v2=
Impulse=Change in momentum =m(v2−v1)=m(v2-v1) …(i)
Again impulse=Area between the graph and time axis
=12×2×4+2×4+12(4+2.5)×0.5+2×2.5=12×2×4+2×4+12(4+2.5)×0.5+2×2.5
=4+8+1.625+5=18.625=4+8+1.625+5=18.625
From (i) and (ii), m(v2−v1)=18.625m(v2-v1)=18.625
⇒v2=18.625m=v1=18.6252+5=14.25m/s⇒v2=
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