Question

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

Answer 1

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