A 2.40 kg block is in equilibrium on an incline of 36.3� degrees.
The acceleration of gravity is 9.81 m/s2 .
What is Fn of the incline on the block?
Answer in units of N
it would be 2.40*9.81 which would get you 23.554N.
oops i also forgot to do 23.554cos(36.3) which would be 18.983.
To find the normal force (Fn) exerted by the incline on the block, we need to analyze the forces acting on the block.
Given:
Mass of the block (m) = 2.40 kg
Incline angle (θ) = 36.3 degrees
Acceleration due to gravity (g) = 9.81 m/s^2
First, we need to resolve the force of gravity into components parallel and perpendicular to the incline.
Force of gravity acting on the block (mg):
mg = mass * acceleration due to gravity
mg = 2.40 kg * 9.81 m/s^2
mg = 23.544 N (This is the magnitude of the force of gravity)
Components of the force of gravity:
Perpendicular component (mg * cos θ)
Parallel component (mg * sin θ)
Now, since the block is in equilibrium, the perpendicular component of the force of gravity will be balanced by the normal force (Fn) exerted by the incline.
Fn = mg * cos θ
Fn = 23.544 N * cos 36.3 degrees
Fn ≈ 19.144 N
Hence, the normal force (Fn) exerted by the incline on the block is approximately 19.144 N.
To calculate the normal force (Fn) of the incline on the block, we need to consider the forces acting on the block in equilibrium. In this case, there are two forces: the force of gravity (mg) acting downward and the normal force (Fn) acting perpendicular to the incline.
First, let's find the force of gravity. The force of gravity is given by the formula:
Fg = mg
where m is the mass of the block (2.40 kg) and g is the acceleration due to gravity (9.81 m/s^2).
Fg = (2.40 kg) x (9.81 m/s^2)
Fg = 23.544 N
Next, we need to find the component of the force of gravity that acts perpendicular to the incline. We can do this by multiplying the force of gravity by the cosine of the angle of the incline (36.3 degrees):
Fn = Fg * cos(36.3 degrees)
Fn = 23.544 N * cos(36.3 degrees)
Fn ≈ 19.200 N
Therefore, the normal force (Fn) of the incline on the block is approximately 19.200 N.