(a) The mass of the block is 5.1 kg.
(b) The magnitude of the normal force of the block is 46.98 N.
(c) The maximum static frictional force between the block and the incline is 23.49 N.
(d) The gravitational force parallel to the plane is 17.1 N.
(e) The block will not slide after being placed on the incline since the parallel force is less than the maximum static frictional force.
(f) The acceleration will be in the direction of the static frictional force.
The given parameters in the question;
- Weight of the metal block, W = 50 N
- Inclination of the plane, θ = 20⁰
- Coefficient of static friction, μs = 0.5
- Coefficient of kinetic friction, μk = 0.3
Mass of the block
The mass of the block is calculated as follows;
W = mg
where;
- g is acceleration due to gravity
m = W/g
m = (50) / 9.8
m = 5.1 kg
Magnitude of the normal force of the block
Fₙ = mg cosθ
Fₙ = W cosθ
Fₙ = 50 x cos(20)
Fₙ = 46.98 N
Maximum static frictional force between the block and the incline
Fs = μs(Fₙ)
Fs = 0.5 x 46.98 N
Fs = 23.49 N
Gravitational force parallel to the plane
F = W sinθ
F = 50 x sin(20)
F = 17.1 N
Net force on the metal block
F(net) = F - Fs
F(net) = 17.1 N - 23.49 N = -6.39 N
Thus, the block will not slide after being placed on the incline since the parallel force is less than the maximum static frictional force.
Acceleration of the block after being placed on the incline
a = F(net)/m
a = -6.39/5.1
a = -1.25 m/s²
The acceleration will be in the direction of the static frictional force.
Learn more about frictional force here: https://brainly.com/question/4618599
#SPJ1
