A mass m = 15.0 kg rests on a frictionless table and accelerated by a spring wit

Question

A mass m = 15.0 kg rests on a frictionless table and accelerated by a spring with spring constant k = 4920.0 N/m. The floor is frictionless except for a rough patch. For this rough path, the coefficient of friction is ?k = 0.5. The mass leaves the spring at a speed v = 3.8 m/s.
1) How much work is done by the spring as it accelerates the mass?
108.3 J

2) How far was the spring stretched from its unstreched length?
0.21 m

3) The mass is measured to leave the rough spot with a final speed vf = 2.1 m/s.
How much work is done by friction as the mass crosses the rough spot?
?????

4) What is the length of the rough spot?
1.02m

5) In a new scenario, the block only makes it (exactly) half-way through the rough spot. How far was the spring compressed from its unstretched length?
???

6) In this new scenario, what would the coefficient of friction of the rough patch need to be changed to in order for the block to just barely make it through the rough patch?
????

Explanation / Answer

This problem has 2 parts. First, it is a problem of conservation of momentum, which you solved already for the initial velocity of the bullet/block compound system at 0.9337701849 m/s. The second part deals with the conservation of energy. The relevant equation states that the initial kinetic energy of the bullet/block is equal to the potential energy the spring stores at maximum deflection, (or, amplitude). The expression for kinetic energy, E, of the bullet/block is E = 1/2 MV^2, where M is the mass of the bullet/block and V is the velocity of it, which you calculated. The expression for the potential energy, P, stored by the spring at amplitude is 1/2 kx^2 where k is the spring coefficient, and x is the amplitude. Since these two expressions are equal for this problem, 1/2 MV^2 = 1/2 kx^2 Substituting the known values, 1/2 (6409.5 g)(0.9337701849 m/s)^2 = 1/2 (6000 n/m)x^2 solving for x, 2.79430728 g*m^2/(s^2) = 3000000 g*(s^2) (x^2) 0.00000093143576 m^2 = x^2 0.000965m = x so the amplitude of the spring's oscillation after impact is 0.000965m, or 0.965mm.

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