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6.2 A body of mass 2 kg initially at restmoves under the action of an appliedhorizontal force of 7 N on a table withcoefficient of kinetic friction = 0.1.Compute the(a) work done by the applied force in10 s.(b) work done by friction in 10 s.(c) work done by the net force on thebody in 10 s.(d) change in kinetic energy of thebody in 10 s,​

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Answer:

Here, u=0(INITIAL speed)

Here, u=0(initial speed)Friction force=0.1*2*10=2N

Here, u=0(initial speed)Friction force=0.1*2*10=2N Net force on the body=7N-2N=5N

Here, u=0(initial speed)Friction force=0.1*2*10=2N Net force on the body=7N-2N=5NSo,a=5/2kg=2.5m/s

Here, u=0(initial speed)Friction force=0.1*2*10=2N Net force on the body=7N-2N=5NSo,a=5/2kg=2.5m/s 2

Here, u=0(initial speed)Friction force=0.1*2*10=2N Net force on the body=7N-2N=5NSo,a=5/2kg=2.5m/s 2

Here, u=0(initial speed)Friction force=0.1*2*10=2N Net force on the body=7N-2N=5NSo,a=5/2kg=2.5m/s 2 Therefore, distance travelled in 10 SEC =1/2at

Here, u=0(initial speed)Friction force=0.1*2*10=2N Net force on the body=7N-2N=5NSo,a=5/2kg=2.5m/s 2 Therefore, distance travelled in 10 sec =1/2at 2

Here, u=0(initial speed)Friction force=0.1*2*10=2N Net force on the body=7N-2N=5NSo,a=5/2kg=2.5m/s 2 Therefore, distance travelled in 10 sec =1/2at 2

Here, u=0(initial speed)Friction force=0.1*2*10=2N Net force on the body=7N-2N=5NSo,a=5/2kg=2.5m/s 2 Therefore, distance travelled in 10 sec =1/2at 2 =1/2∗2.5∗100

Here, u=0(initial speed)Friction force=0.1*2*10=2N Net force on the body=7N-2N=5NSo,a=5/2kg=2.5m/s 2 Therefore, distance travelled in 10 sec =1/2at 2 =1/2∗2.5∗100 =125 m

Here, u=0(initial speed)Friction force=0.1*2*10=2N Net force on the body=7N-2N=5NSo,a=5/2kg=2.5m/s 2 Therefore, distance travelled in 10 sec =1/2at 2 =1/2∗2.5∗100 =125 mNow, WORK done by the applied force=F.S =125*7=875 J


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