block sliding please help

Question

A mass m = 14 kg rests on a frictionless table and accelerated by a spring with spring constant k = 4983 N/m. The floor is frictionless except for a rough patch. For this rough path, the coefficient of friction is μ_k = 0.53. The mass leaves the spring at a speed v = 3 m/s.

1) What is the length of the rough spot?

2) 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?

3) 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?

Answer 1

1. If the block was sliding on a frictionless surface, it would never come to a stop. We have to assume that the rough patch is rough enough to bring the block to a stop at the end of the patch. As the patch has friction, it gives a retarding force to the object's motion. In this case the equation μ_kN = -Ma applies where N is the normal force (14 kg x 9.8 m/s²). Putting in the given numbers, we get 0.53 x 14 x 9.8 = 14 x a. Solving for a, we get -5.19 m/s². To find the distance to stop (the length of the patch), we can use the equation V² = V₀² + 2aS. As a is negative and V =0, the S = -V_o² / -2a. Plugging in the numbers we get -3²/-2 x 5.19 = 0.87 m as the length of the patch.
2. New distance on patch = 0.87/2 = 0.435 m. Using V² = V₀² + 2aS again, V² =0 then V_o² = 2x5.19x0.435. Therefore, the new starting velocity is √ 2x5.19x0.435 = 2.12 m/s. To find the amount the spring was stretched, it is necessary to recall that the kinetic energy given to the mass is equal to the stored potential energy of the spring. The relationship is 1/2MV² = 1/2kX² where X is the displacement of the spring from its un-stretched length. V in this case is 2.12 m/s already calculated. Solving for X, we get 0.077 m.
3. For this scenario, we need to calculate the new deceleration on the patch as the coefficient of friction has changed. We again make use of V² = V₀² + 2aS, this time solving for a, knowing that V_o = 2.12 m/s, S = 0.87 m, V = 0. a = -2.58 m/s². Now set up the equation, μN = Ma and solve for μ after substituting N, M, and new a. I got 0.26.

Please comment if and when you know the correct answers. I like to know if I got it right. Thanks.