A massless spring of constant k = 76.5 N/m is fixed on the left side of a level

Question

A massless spring of constant k = 76.5 N/m is fixed on the left side of a level track. A block of mass m = 0.50 kg is pressed against the spring and compresses it a distance of d. The block (initially at rest) is then released and travels toward a circular loop-the-loop of radius R = 1.5 m. The entire track and the loop-the-loop are frictionless, except for the section of track between points A and B. Given that the coefficient of kinetic friction between the block and the track along AB is µk = 0.34, and that the length of AB is 2.5 m, determine the minimum compression d of the spring that enables the block to just make it through the loop-the-loop at point C. [Hint: The force of the track on the block will be zero if the block barely makes it through the-loop-the-loop.]

Explanation / Answer

let v is required minimum speed at the top of the loop,

Apply, Fnet = m*a_rad

m*g = m*v^2/R

==> v = sqrt(g*R)

= sqrt(9.8*1.5)

= 3.83 m/s

let x is compression of te spring.

Now use Work-energy throrem,

Net workdone = change in kinetic energy

W_spring + W_friction + W_gravity = 0.5*m**v^2

0.5*k*x^2 + mue_k*m*g*d*cos(180) + m*g*(2*R)*cos(180) = 0.5*m*v^2

0.5*76.5*x^2 - 0.34*0.5*9.8*2.5 - 0.5*9.8*2*1.5 = 0.5*0.5*3.83^2

38.25*x^2 = 0.34*0.5*9.8*2.5 + 0.5*9.8*2*1.5 + 0.5*0.5*3.83^2

38.25*x^2 = 22.53

x^2 = 22.53/38.25

x^2 = 0.589

x = sqrt(0.589)

= 0.767 m <<<<<<<<<<<<<---------------------------------Answer

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