Question

6.
Two blocks P and Q of masses 5 kg and 10 kg
respectively are placed on a rough horizontal surface
and a horizontal force of 25 N is applied on the block
Pas shown in figure. The frictional force exerted by
the surface on the block Q will be (g = 10 m/s2)
F = 25
N
P
(us = 0.4, Mk = 0.3) (us = 0.5, Mk = 0.4)
(1) 10 N
(2) 5N
(3) 50 N
(4) 40 N
A one of length 5 m has uniform mass per unit​

Answer 1

Answer:

The frictional force exerted by the surface on the block Q will be 5 N

Explanation:

Total limiting friction or maximum possible friction between the blocks and the ground surface is given as

$F_f = \mu_s(m_1 + m_2)g$

$F_f = 0.4(5)10 + 0.5(10)10$

$F_f = 20 + 50 N$

so here the system of two block will not slide when applied force is less than 70 N

Now limiting friction on block P

$F_{f1} = \mu_s m_1 g$

$F_{f1} = 0.4(5)(10)$

$F_{f1} = 20N$

So here when we apply 25 N force on P block then the net force exerted by it on Q block is given as

$F = 25 - 20$

$F = 5 N$

now the force exerted by the ground as friction force is given as

[tex]F_f = 5 N

#Learn

Topic : Friction

https://brainly.in/question/4582605

Answer 2

Explanation:

find the attachment below

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