Question

A wooden box kept on a rough horizontal surface is pushed by Jai and Samarth from the same side in such a way that the force applied by Samarth is twice that applied by Jai. The frictional force acting between the surface of the box and the ground is one-tenth of the total force applied by Jai and Samarth. As a result, the box experiences a net unbalanced force of 27 N. What is the force applied by Samarth?
(1) 15 N
(2) 20 N
(3) 27 N
(4) 10 N

Answer 1

since both rare in same direction forces will be added .friction force is in other direction so it will be subtracted
x+2x-1÷10×3x=27
3x+3x÷10 =27
30x-3x=270
27x=270
x=10
therefore force by samarth =2×x
20N

Answer 2

Answer:

Step-by-step explanation:

given data:

(1)  Force  by Smarath = 2 * Force(F) by Jai

(2) Force due to friction = 1/10th of (total force of S & J)

(3) Net force on wooden block = 27N

what we need to find is how much force Samarth is applying alone ?

## REQUIRED CONCEPTUAL KNOWLEDGE TO SOLVE THIS PROBLEM IS:

frictional force is always opposite to direction of  external force .

As it is given in the question that force applied by Samarath and Jai is in same direction therefore force will be additive in nature.

SO total force applied by them = [ F + 2F ] = 3F   (where F is force applied by Jai)

therefore according to given info frictional force will be =

                                                                     1/10 of 3F = (3/10)*F

now net force on block-------------= (3F - frictional force) = 27N

                                                                     3F - (3/10)*F = 27N

                                                                           (27/10)*F =27N

                                                                                       F = 10N

ANS: force applied by samarth is 2F = 2*10 = 20N