A massless spring of constant k = 75.0 N/m is fixed on the left side of a level

Question

A massless spring of constant k = 75.0 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, as in the figure shown below. 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 = 0.26, 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 exerted by the track on the block will be zero if the block barely makes it through the-loop-the-loop.

* Answer must be in "meters"

Explanation / Answer

Potential Energy at the top the loop = mgh = 0.5 * 9.8 * 2R = 0.5 * 9.8 * 2 * 1.5 = 14.7 joules
Kinetic energy at the bottom B should be equal to this value so that ball reaches C
Kinetic energy at B = 14.7 joules

energy dissipated due to friction = uN = umg = 0.26 * 0.5 * 9.8 =1.274 joules (u= coefficient of friction, N =normal force being equal to the weight of the mass)

so total energy that has to be at A and the during when the spring was compressed = 14.7 joules + 1.274 joules = 15.974 joules

and this should be equal to the energy stored in spring during its compression so that it can be delivered to the block
so spring energy = 15.974 = (1/2) * k * d^2 (where k =spring constant = 75 and only unknown is d )

on solving above equation we get
d= 0.65 m (answer)

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