Mass of particle , m = 0.3 kg .
Speed of object at position A , [tex]v_a=1 \ m/s.[/tex]
Kinetic energy of object at position B , [tex]K_b=7.6\ J.[/tex]
Now,
A) Kinetic energy of object at position A , [tex]K_a=\dfrac{1}{2}mv_a^2=\dfrac{1}{2}\times 0.3\times 1^2=0.15\ J.[/tex]
B) We know,
[tex]K_b=\dfrac{1}{2}mv_b^2\\\\v_b=\sqrt{\dfrac{2K_b}{m}}[/tex]
Putting all values in above equation :
We get ,
[tex]v_b=\sqrt{\dfrac{2K_b}{m}}=\sqrt{\dfrac{2\times 7.6}{0.3}}=7.12\ m/s.[/tex]
C) We know by work energy theorem :
Work Done = Change in Kinetic Energy
[tex]W=K.E_b-K.E_a\\\\W=7.60-0.15\ J=7.45\ J.[/tex]
Hence , this is the required solution .
Learn More :
Work and Energy
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