Answer:
=====
Explanation:
a) ForceDownPlane = mg sin Φ
= 2 kg * 9.81 m/s^2 sin 30 = 9.81 N
b) F = ma
F = 2 kg (2.3 m/s^2) = 4.6 N
this is the net force accelerating the block
ForceDownPlane - Ffriction = 4.6 N
9.81 - 4.6 = 5.2 N of friction force
c) coeff of kinteic friction
normal force = mg cos 30
normal force * coefficient = 5.2
coefficient of kinteic friction = .31
( NOTE: UNITS of acceleration are m/ s^2 )
Answer:
a) 9.8N
b) 5.2 N
c) 0.306
Explanation:
Given:
a. Component of the block's weight along the ramp:
The component of the weight along the ramp ([tex] F_{\textsf{parallel}} [/tex]) can be found using the formula:
[tex] F_{\textsf{parallel}} = mg \sin(\theta) [/tex]
[tex] F_{\textsf{parallel}} = 2.0 \times 9.8 \times \sin(30^\circ) [/tex]
[tex] F_{\textsf{parallel}} = 2.0 \times 9.8 \times 0.5 [/tex]
[tex] F_{\textsf{parallel}} = 9.8 \, \textsf{N} [/tex]
So, the component of the block's weight along the ramp is [tex] 9.8 [/tex] N.
b. Frictional force between the block and the ramp:
The net force ([tex] F_{\textsf{net}} [/tex]) acting along the ramp can be related to the acceleration ([tex] a [/tex]) using Newton's second law:
[tex] F_{\textsf{net}} = m \cdot a [/tex]
The net force is the difference between the component of the weight along the ramp and the frictional force ([tex] f [/tex]):
[tex] F_{\textsf{net}} = F_{\textsf{parallel}} - f [/tex]
Substitute in the values:
[tex] 2.0 \cdot 2.3 = 9.8 - f [/tex]
[tex] f = 9.8 - 4.6 [/tex]
[tex] f = 5.2 \, \textsf{N} [/tex]
So, the frictional force between the block and the ramp is [tex] 5.2 [/tex] N.
c. Coefficient of kinetic friction ([tex] \mu_k [/tex]):
The coefficient of kinetic friction can be found using the formula:
[tex] \mu_k = \dfrac{f}{F_{\textsf{normal}}} [/tex]
The normal force ([tex] F_{\textsf{normal}} [/tex]) is equal to the component of the weight perpendicular to the ramp:
[tex] F_{\textsf{normal}} = mg \cos(\theta) [/tex]
[tex] F_{\textsf{normal}} = 2.0 \times 9.8 \times \cos(30^\circ) [/tex]
[tex] F_{\textsf{normal}} = 2.0 \times 9.8 \times \dfrac{\sqrt{3}}{2} [/tex]
[tex] F_{\textsf{normal}} = 2.0 \times 9.8 \times 0.866 [/tex]
[tex] F_{\textsf{normal}} \approx 16.97409791 \, \textsf{N} [/tex]
Now, calculate [tex] \mu_k [/tex]:
[tex] \mu_k = \dfrac{5.2}{16.97409791} [/tex]
[tex] \mu_k \approx 0.3063491224 [/tex]
So, the coefficient of kinetic friction is approximately [tex] 0.306 [/tex].
