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
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
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,
To find the frictional force acting on the block, we need to first calculate the normal force and then use it to determine the frictional force.
1. Calculate the vertical component of the applied force:
The vertical component of the applied force is given by F_vertical = F * sin(θ), where F is the applied force (20 N) and θ is the angle at which the force is applied (25°).
F_vertical = 20 N * sin(25°) ≈ 8.68 N
2. Calculate the normal force:
The normal force (N) is equal to the weight of the block (mg), where m is the mass of the block (3 kg) and g is the acceleration due to gravity (10 m/s²).
N = m * g = 3 kg * 10 m/s² = 30 N
3. Calculate the frictional force:
The frictional force (F_friction) can be determined using the equation F_friction = μ * N, where μ is the coefficient of kinetic friction (0.3) and N is the normal force.
F_friction = 0.3 * 30 N = 9 N
Therefore, the frictional force acting on the block is most nearly 9 N.