A block of mass M = 3.1 kg is at rest on an incline (theta = 28degree) that has

Question

A block of mass M = 3.1 kg is at rest on an incline (theta = 28degree) that has friction between the incline's surface and the base of the block. The block sits a distance L from the bottom of the incline. Attached to the block is a stretched spring (i.e., L > Xo ) that has a spring constant k = 24 N/m and a natural length of Xo = 2.6 meters. Suppose the magnitude of the frictional force acting on the block is Ffr =18.3 Newtons. Use g = 9.81 m/s2. What is Fs , the magnitude of the spring force acting on the block? What is L , the total length of the spring? Given that the coefficient of static friction between the incline and the block is mu s = 0.83, what is the maximum distance, Lmax , from the bottom of the incline the block can be so that it is still at rest? Suppose theta, the angle of the incline is decreased to theta = 14degree. How does Lmax, the maximum length of the spring to keep the block at rest, change, if at all? Lmax decreases Lmax increases Lmax stays the same

Explanation / Answer

sum forces parallel to the slope
mg sin theta + Fs - Fr = 0

3.1*9.81*sin(28) + F - 18.3 = 0

F=4.023 N

b) F = k (L -x0)

4.023 = 24*(L-2.6)
L=2.77 m

c)
3.1*9.81*sin(28) + 24*(L-2.6) - 0.83*3.1*9.81*cos(28)=0

L=2.934 m

d)

3.1*9.81*sin(14) + 24*(L-2.6) - 0.83*3.1*9.81*cos(14)=0
L=3.31

gets bigger

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