Answer:
Explanation:
1 ) When there is no friction , gravitational force on the block = mgsin40
acceleration = gsin40
= 6.3 m /s²
2 )the acceleration of the block after it begins to slide down the plane
Net force on the block
= mgsin40 - frictional force
mgsin40 - μk mgcosθ
= mg ( sin40 - .38 x cos40)
acceleration
= g ( sin40 - .38 x cos40)
9.8 ( sin40 - .38 x cos40)
= 9.8 ( .6428 - .29)
= 3.46 m /s²
3 ) Net downward force
= 8.8 x 3.46
= 30.448 N
To balance it
restoring force by spring
k x = 30.448
k = 30.448 / .13
= 234.21 N/m
4 )
for equilibrium
mgsin40 = μmgcos40
μ = sin40 /cos40
= tan40
= .84
(1) When there is no friction, the magnitude of the acceleration of the block is [tex]6.29\,m/s^2[/tex].
(2) With friction, the magnitude of the acceleration of the block after it begins to slide down the plane is [tex]3.45\,m/s^2[/tex].
(3) The minimum spring constant of the spring to keep the block from sliding is 3.94 N/m.
(4) Minimum value for the coefficient of static friction needed between the new block and the plane to keep the system from accelerating is 0.24.
(1) In the absence of friction, the weight of the block (gravitational force) is the only force on the block on the block
[tex]W= m_1 g\,sin\, \theta[/tex]
i.e., acceleration is given by;
[tex]a=g\,sin\, \theta[/tex]
But given that, [tex]\theta = 40^\circ[/tex]
[tex]a = g \,sin\, 40^\circ= 9.8\,m/s^2 \times 0.6427=6.29\,m/s^2[/tex]
(2) When the block begins to slide down the inclined plane the net force can be obtained using Newton's second law of motion.
[tex]\sum F = ma[/tex]
From the free body diagram, the net force along the horizontal direction, we get;
[tex]\sum F =mg\,sin\,40^\circ - \mu_k\,N[/tex]
But, [tex]N = mg\, cos \,40^\circ[/tex]
[tex]\therefore\,\sum F =mg\,sin\,40^\circ - \mu_k\,mg\,cos\,40^\circ= ma_{net}[/tex]
[tex]\implies a_{net}=g\,sin\,40^\circ - \mu_k\,g\,cos\,40^\circ[/tex]
[tex]\implies a_{net}=(9.8m/s^2 \times 0.6427)-(0.38\times 9.8\,m/s^2 \times 0.766)=3.45 \, m/s^2[/tex]
(3) The net downward force can be given by;
[tex]F_{net}=ma_{net}=(8.8\,kg) \times (3.45\,m/s^2)=30.36\,N[/tex]
Therefore, the spring must exert a force equal to this net force to keep the mass from accelerating. i.e.;
[tex]F = kx= 30.36\,N[/tex]
[tex]\implies k=\frac{F}{x}= \frac{30.36\,N}{0.13\,m} =3.94\,N/m[/tex]
4 ) In order to keep the system from accelerating, the frictional force due to the second block must be equal to the net horizontal force of the first block.
[tex]\mu_km_2 g\,cos\, \theta= F_{net}[/tex]
[tex]\implies \mu_k = \frac{F_{net}}{m_2 g\,cos\,40}=0.24\,[/tex]
Learn more about Newton's second law of motion here:
https://brainly.com/question/13881699
