Question

A time dependent force F = 1012 starts acting on a block of mass 10kg kept at rest on a rough horizontal surface att 0.1
the coefficient of static friction between block and surface is 0.38 and that of kinetic friction is 0,35. What is the value of
frictional force acting on the block att = 25? (g = 10ms?)​

Answer 1

Answer:

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