Two block of masses 3kg and 2kg are placed in contact with each other on a frictionless table. Find then force on the common cross-sectional area of contact if a force of 5N is applied on bigger block smaller block.
DEADPOOL1111:
first we need to find acceleration.acceleration = net force /total mass so 5/(3+2) =1m/s^2.then draw free body diagram of the two bodies together with 2kg block first and 3kg block 2nd.draw acceleration from 2kg block to 3kgblock.then F = m x a .so force applied by 2kg on 3kg is F = 2Kg x 1 m/s^2.so force,F = 2N
first we need to find acceleration.acceleration = net force /total mass so 5/(3+2) =1m/s^2.then draw free body diagram of the two bodies together with 2kg block first and 3kg block 2nd.draw acceleration from 2kg block to 3kgblock.then F = m x a .so force applied by 2kg on 3kg is F = 2Kg x 1 m/s^2.so force,F = 2N
Answers
Answered by
2
1St one is bigger
as it required more force than other
as it required more force than other
plzzzz explain in details...
Answered by
4
the force applied by 2kg block on 3kg block is 2N.
Similar questions