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
Explanation:
If the center of the load is directly above the vertebrae, there is no torque in the system. This is a good thing so that the vertebrae are not put out of alignment over time. (Of course, this still doesn't prevent compression of the vertebrae over time, which is a possibility.)
<em><u>1.car</u></em><em><u> </u></em><em><u>towing</u></em>
<em><u>2.pulling</u></em><em><u> </u></em><em><u>bucket</u></em><em><u> </u></em><em><u>of</u></em><em><u> </u></em><em><u>water</u></em>
<em><u>3.gym</u></em><em><u> </u></em><em><u>equipment</u></em><em><u> </u></em>
<em><u>4.crane</u></em><em><u> </u></em><em><u>machine</u></em>
<em><u>5.tug</u></em><em><u> </u></em><em><u>of</u></em><em><u> </u></em><em><u>war</u></em>
Either it don't have a magnetic field around them or it's not a sheet steel door