- #1
Ry122
- 565
- 2
A 23.0 kg block is initially at rest on a horizontal surface. A horizontal force of 76.0 N is required to set the block in motion. After it is in motion, a horizontal force of 60.0 N is required to keep the block moving with constant speed. Find the coefficients kinetic friction from this information.
my attempt:
After the block is in motion 60N is required to prevent it from accelerating.
Therefore what is left over must be used to overcome friction which is 16N.
Friction force = weight x coefficient of kinetic friction
16N=23x9.8 x ck
ck=.070
But this is incorrect.
What am i doing wrong?
my attempt:
After the block is in motion 60N is required to prevent it from accelerating.
Therefore what is left over must be used to overcome friction which is 16N.
Friction force = weight x coefficient of kinetic friction
16N=23x9.8 x ck
ck=.070
But this is incorrect.
What am i doing wrong?