a block of mass 10 kg is placed on smooth horizontal surface. five equal forces each of 10N is applied to the block. if after every second one force is removed then the distance covered by block in 8 seconds is __________
Answers
Answered by
3
Answer:
for t=0−4s
F=
4
100
t=25t
maximum friction force f
max
=0.5×100=50N
at t = 2 sec F
ext
=.f
max
F= 50 N so motion of block starts
at t = 4 s
acceleration from 2s to 4 s is given as a=
10
25t−50
t=2.5t−5
velocity is area under a-t diagram
v=∫
2
4
adt
2
1
(4−2)×5=5m/s
v=5m/s at t = 4 s (maximum velocity)
From t=4-7 s Kinetic friction will be acting as block is in motion a=
m
F−fmax
=
10
40−50
=−1m/s
2
at t=7
v
′
=5−1×(7−4)
v
′
=2m/s
after 7 s the force is removed acceleration is given as a=
m
F−fmax
=
10
0−50
=−5m/s
2
The block will come to rest after t s
0=2−5(t−7)⇒t=7.4s
Time of motion =7.4−2=5.4s
at, t=1 F=25
sof=25 at∑F=0
Answered by
1
ANS=33N.............
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