Question

Two identical 5kg blocks are moving with same speed of 2m/s towards each other along a frictionless horizontal surface . The blocks collide , stick together and comes to rest . Considering the two blocks as a system , calculate the work done by (1) external forces (2) internal forces

Answer 1

@akash123

Answer 2

work done by all the forces= change in the kinetic energy

Answer 3

okay

Answer 4

so what are the external forces exerting on these two blocks?

Answer 5

i did'n get the question , what external and what internal ?

Answer 6

gravity ?

Answer 7

there is no friction...so only external forces are gravity and normal reaction

Answer 8

okay

Answer 9

internal forces..when they'll collide with each other then they will exert forces on each other..that's the internal force

Answer 10

okay

Answer 11

it's fine?

Answer 12

ok

Answer 13

now will the external forces( gravity and normal reaction) do work?

Answer 14

no ?

Answer 15

yes...they wont do any work because they are perpendicular to d displacement...so no work done

Answer 16

okay understood

Answer 17

so from work energy theorem
work done= change in the kinetic energy
--> work done by external forces+ work done by the internal forces= change in the kinetic energy of the system

Answer 18

so u can calculate the work done by the internal forces

Answer 19

u know the work done by the external forces(=0) and the change in the KE of the system

Answer 20

okay

Answer 21

u'll calculate the change in the KE of the system?

Answer 22

final KE=0
initial KE= 1/2 m1 v1^2 + 1/2 m2 v2^2

Answer 23

okay

Answer 24

m1= m2=5kg
and v1=v2=2m/s

Answer 25

20 J ?

Answer 26

yes

Answer 27

so work done by internal forces= (final KE- initial KE)= -20 J

Answer 28

okay

Answer 29

u got the meaning of -ve sign? I mean...why work done is -ve?

Answer 30

yes 0-20 ?