A block with mass m = 14 kg rests on a frictionless table and is accelerated by

Question

A block with mass m = 14 kg rests on a frictionless table and is accelerated by a spring with spring constant k = 4253 N/m after being compressed a distance x1 = 0.531 m from the spring’s unstretched length. The floor is frictionless except for a rough patch a distance d = 2.8 m long. For this rough path, the coefficient of friction is k = 0.45

Instead, the spring is only compressed a distance x2 = .131 m before being released. How far into the rough path does the block slide before coming to a rest?

Explanation / Answer

Use conservation of energy to determine the energy of the block as it leaves the spring. The potential energy of the spring becomes the kinetic energy of the block.

As for a spring, F=kx, where F is force, k is the spring constant, and x is the distance the spring is stressed or compressed from equilibrium, U (potential energy) = 0.5*k(x2), as energy is the integral of velocity with respect to distance (x). K is kinetic energy.

K=U=0.5*k*x2
K=0.5*(4253 N/m)*(0.131 m)2 = 36.492 J

We can use this energy value to determine how far the block will go as the friction will "consume" all of this kinetic energy.

Now as you know, the magnitude of the force of kinetic friction on a flat surface is equivalent to...

F=mg

Now the integral of force with respect to displacement (or force * displacement) is work, so...

W=mgy

Where y is your answer...so if we set K=W...

K=mgy
K/(mg)=y
(36.492 J)/((0.45)*(14 kg)*(9.8 m/s^2)) = y = 0.591 m

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