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
784 Newtons or 176.37 lbs
Answer:
0.74 N/cm
Explanation:
The following data were obtained from the question:
Mass (m) = 3 Kg
Extention (e) = 40 cm
Spring constant (K) =?
Next, we shall determine the force exerted on the spring.
This can be obtained as follow:
Mass (m) = 3 Kg
Acceleration due to gravity (g) = 9.8 m/s²
Force (F) =?
F = mg
F = 3 × 9.8
F = 29.4 N
Finally, we shall determine the spring constant of the spring. This can be obtained as follow:
Extention (e) = 40 cm
Force (F) = 29.4 N
Spring constant (K) =?
F = Ke
29.4 = K × 40
Divide both side by 40
K = 29.4 / 40
K = 0.74 N/cm
Therefore, the spring constant of the spring is 0.74 N/cm
Answer:
48.6°
Explanation:
The forward force, F equals the component of the weight along the slope.
So mgsinθ = ma where a = acceleration and θ = angle between the slope and the horizontal.
So a = gsinθ
Since we are given that a = 75%g = 0.75g,
0.75g = gsinθ
sinθ = 0.75
θ = sin⁻¹(0.75)
= 48.6°
