A body of mass 5kg is moving with a momentum of 10kgms-1.A force of 0.2N acts on it in direction of motion of body for 10sec.The increase in kinetic energy is?

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Answer 1 · 2018-05-11T16:05:00

#"4.4 J"#

Initial velocity of body is

#"v"_i = "p"/"m" = (10 cancel"kg" "m/s")/(5 cancel"kg") = "2 m/s"#

Acceleration of body when force is applied is

#"a" = "F"/"m" = "0.2 N"/"5 kg" = "0.04 m/s"^2#

Velocity of body after #10\ "s"# is

#"v"_f = "v"_i + "at"#

#"v"_f = "2 m/s + (0.04 m/s"^2 × "10 s) = 2.4 m/s"#

Increase in kinetic energy

#"ΔK" = 1/2"m(v"_f^2 - "v"_i^2")"#

#"ΔK" = 1/2 × "5 kg" × [("2.4 m/s")^2 - ("2 m/s")^2] = "4.4 J"#