Question

· Using a horizontal force of 200 N, we intend to move a wooden
cabinet across a floor at a constant velocity. What is the friction
force that will be exerted on the cabinet?​

Answer 1

Answer:

200 N

Step-by-step explanation:

Since the cabinet moves with constant velocity, its acceleration will be zero and the net force acting on it should be zero.

Hence frictional force acting directly opposite to the direction of the applied force to nullify the effect of the applied force will be zero, making the cabinet move with constant velocity.

$F_{net}$ = F - $F_{k}$

Thus,

Forth on cabinet - Friction = 0

200 N - $F_{k}$ = 0

$F_{k}$ = 200 N - 0

$F_{k}$  = 200 N

Answer 2

GIVEN : F = 200 N

---> VELOCITY IS CONSTANT UP TO WHEN THE FORCE IS APPLIED ON THE WOODEN CABINET.SO , AT ANY INSTANT IN WHICH THE FORCE IS APPLIED, THE VELOCITY WILL ONLY BE ' U ' THAT IS CONSTANT VELOCITY.

TO FIND : FRICTION FORCE EXERTED ON THE CABINET.

EXPLANATION :

--> FRICTION FORCE IS THE FORCE THAT ACTS OPPOSITE TO THE APPLIED FORCE.

--> SO THE NET FORCE ON THE OBJECT WILL BE APPLIED FORCE MINUS ( - ) FRICTION FORCE.

--> NET F. = APPLIED F. - FRICTION F.

--> NET F. = 200 N - FRICTION F.

--> FRICTION F. = 200 N - NET F.

--> FRICTION F. = 200 N - Ma [ F =Ma ]

--> FRICTION F. = 200 N - M ( V - U )

t

[ a = v - u ]

t

--> FRICTION F. = 200 N - M ( U - U )

t

--> HERE U = V , BECAUSE VELOCITY IS CONSTANT.

--> FRICTION F. = 200 N - M ( 0 )

t

--> FRICTION F. = 200N - 0

--> FRICTION F. = 200N

--> HENCE THE FRICTION FORCE IS 200N.

HOPE THIS HELPS!