A body of mass 5kg is moving with a momentum of 10 kgm/s. A force of 0.2N acts on it in the direction of motion of the body for 10 sec. The increase in its kinetic energy is?

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Dec 13, 2016

Answer:

#"4.4 J"#

Explanation:

Before the force is applied we know

#"momentum" = "mass" times "velocity" = "10 kg m/s"#

Since mass in the above is #"5 kg"#, then velocity = #"2 ms"^(-1)#

We can now work out kinetic energy, KE:

#KE=1/2mv^2=1/2*5*2^2="10 J"#

The force of #"0.2 N"# is now applied in the direction of travel for #10# seconds. Newton's second law can be expressed as "the force applied equals the rate of change of momentum". OR

#F=(mv-"mu")/t=m * (v-u)/t#

Where #v# is the final velocity after applying force, and #u# is the initial velocity ( #"2 ms"^(-1)# in this case from above)

So

#0.2 = 5 * ((v-2))/10#

rearranging helps us find #v#

#2 = 5 * (v-2)#

#2/5 = v-2#

#v = 12/5" ms"^(-1)#

Hence final kinetic energy, #KE#, is

#KE = 1/2mv^2 = 1/2 * 5 * (12/5)^2 = "14.4 J"#

So the overall change in #KE# is

#Delta_"KE" = 14.4 - 10 = "4.4 J"#

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