Respuesta :
Answer:
a) a = 4.9 m / s², N = 16.97 N and b) F = 9.8 N
Explanation:
a) For this exercise we will use Newton's second law, we write a reference system with the x axis parallel to the plane, see attached, in this system the only force we have to break down is weight, let's use trigonometry
sin 30 = Wx / W
cos 30 = Wy / W
Wx = W sin30
Wy = W cos 30
Let's write the equations on each axis
X axis
Wx = ma
Y Axis
N- Wy = 0
N = Wy = mg cos 30
N = 2.0 9.8 cos 30
N = 16.97 N
We calculate the acceleration
a = Wx / m
a = mg sin 30 / m
a = g sin 30
a =9.8 sin 30
a = 4.9 m / s²
b) For the block to move with constant speed, the acceleration must be zero, so the force applied must be equal to the weight component
F -Wx = 0
F = Wx
F = m g sin 30
F = 2.0 9.8 sin 30
F = 9.8 N
The block’s acceleration down the ramp is 4.9m/s²
The force applied upward along and parallel to the ramp would allow the block to move with a constant velocity is 9.8 N
According to Newton's second law;
[tex]F_m = ma[/tex]
Fm is the moving force
m is the mass of the block
a is the acceleration
Since the block acts along the ramp
[tex]F_m = Wsin \theta\\F_m=mgsin \theta[/tex]
Given the following parameters
m = 2.0kg
g = 9.8m/s²
[tex]\theta[/tex] = 30 degrees
Get the required acceleration:
[tex]a=\frac{F_m}{m}\\a=\frac{mgsin \theta}{m}\\a=gsin \theta[/tex]
Substitute the given parameters
[tex]a=9.8sin30^0\\a=9.8(0.5)\\a=4.9m/s^2[/tex]
Hence the block’s acceleration down the ramp is 4.9m/s²
b) The force applied upward along and parallel to the ramp is the moving force along the plane expressed as:
[tex]F_m=Wsin \theta\\F_m=2(9.8)sin30\\F_m=19.6(0.5)\\F_m=9.8N[/tex]
Hence the force applied upward along and parallel to the ramp would allow the block to move with a constant velocity is 9.8 N
Learn more here: https://brainly.com/question/13598868
