Question

. A horizontal force of 1200 g
wt. is applied to a 1500 gram
law block which rests on a
horizontal surface. If the
coefficient of friction is 0.3
Find acceleration produced
in the block (g=980 cm /s2)
[a] 450 cms-2 [b]
490 cms-2
[c] 400
cms-2
(d) 500 cms-2​

Answer 1

Answer:

Horizontal force

F=1.2kgwt

F

1

=1.2×g

Now frictional force

F=umg

F=0.3×1.5×g

F=0.45g

Let acceleration produced in the body be a.

Now we have

F

1

−F=ma

1.2g−0.45g=1.5a

0.75g=1.5a

a=

2

g

a=4.9m/s

2