A mass m = 11.0 kg rests on a frictionless table and accelerated by a spring wit

Question

A mass m = 11.0 kg rests on a frictionless table and accelerated by a spring with spring constant k = 4060.0 N/m. The floor is frictionless except for a rough patch. For this rough path, the coefficient of friction is ?k = 0.52. The mass leaves the spring at a speed v = 3.8 m/s. 1) The mass is measured to leave the rough spot with a final speed vf = 2.0 m/s. How much work is done by friction as the mass crosses the rough spot? 2)What is the length of the rough spot? 3) 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? 4) 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

1)

Wf = K = 0.5 * m * (vi^2 - vf^2) = 0.5*15*(4.1*4.1-2.2*2.2) = 89.775 J

2)

fk = k N = 0.43*15*9.8 = 63.21 N

Wf = fk * d   >>>> d = Wf/fk = 89.775/63.21 = 1.4203 m

3)

fk * (d/2) = 0.5 * k * (A^2)

63.21 * (1.4203/2) = 0.5 * 4578.0 * (A^2)

A = 0.140 m

4)

' = /2 = 0.43/2 = 0.215

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