Question

Two blocks a and b of masses 10 kg and 20 kg respectively are placed on each other in the combination rest on fixed horizontal surface c a light string passing over the smooth light pulley is used to connect a and b as shown the coefficient of sliding friction between all surfaces in contact is new is equals to 1 by 4 if a is direct without force if there for both a and b to move with a uniform speed we have value of f is

Answer 1

Answer:

                The force applied on Block A is 175N

Explanation:

Mass of block A  (M₁)=10kg

Mass of block B  (M₂)=20kg

Co-efficient of friction (μ)=1/4=0.25

As the force applied on block A makes the both blocks A and B move with uniform speed hence the force applied onto it will be equal to the number  of reaction forces:

Formula:

Total Forces (F)=F₁+F₂+.....Fⁿ

Hence, for block A:

F=T=F₁+F₂   (∴ where T=the opposite of F form block B)

F=T+μM₂g+μ(M₁+M₂)g

F=T+μ(M₁+2M₂)g   .......(i)

Force from Block B:

T=F₁=μM₂g         .......(iI)

From Equation (i) and (ii)

F=μM₂g+μ(M₁+2M₂)g

=μ(M₁+3M₂)g

Putting values;    F=1/4(3×20+10)10

F=0.25( 60+10)10

F=175N

Hence, 175N force is applied on Block A.

hope you find it helpful  :)