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
Answers
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:
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