A block of mass 3 kg slides along a horizontal surface while a 20-N force is applied to it at an angle of 25°. If needed, use g = 10 m/s2. For a coefficient of kinetic friction of 0.3 between the block and the surface, the frictional force acting on the block is most nearly
9 N
8.5 N
15 N
6 N
12 N
4 answers
friction=mu*mg=.3*3*10 N
Is the 25 degree angle of the applied force above or below horizontal? It makes a big difference in the answer.
If the angle is 25 degree below the horizontal, then the upward component makes the friction lighter (less normal force).
friction=mu(normal force)
= .3(mg-20sin25)
=.3(30-8.45)
= no answer.
so assuming the 25 is above the horizontal, friction will be greater..
friction= = .3(mg+20sin25)
=.3(38.45)= almost 12 N
friction=mu(normal force)
= .3(mg-20sin25)
=.3(30-8.45)
= no answer.
so assuming the 25 is above the horizontal, friction will be greater..
friction= = .3(mg+20sin25)
=.3(38.45)= almost 12 N
F(y) = F•sinα =20•sin25º =8.45 N.
The force may be applied
\ from top downward
or / from the bottom upward ,
then
F(fr) =μ•N = μ•[mg ± F(y)] =
= 0.3•{3•10 ± 8.45) =
= 11.54 N or 6.5 N,
The force may be applied
\ from top downward
or / from the bottom upward ,
then
F(fr) =μ•N = μ•[mg ± F(y)] =
= 0.3•{3•10 ± 8.45) =
= 11.54 N or 6.5 N,
1 answer