A mass m = 10 kg rests on a frictionless table and accelerated from rest by a sp

Question

A mass m = 10 kg rests on a frictionless table and accelerated from rest by a spring with spring constant k = 4490 N/m. The floor is frictionless except for a rough patch. For this rough patch, the coefficient of the kinetic friction is µk = 0.5. The mass leaves the spring with speed v = 2.8 m/s. Use work and energy to answer the following questions:

(a) How much work is done by the spring as it accelerates the mass? How far was the spring compressed from its unstretched length? (4pt) (b) The mass is measured to leave the rough spot with a final speed vf = 1.3 m/s. How much work is done by friction as the mass crosses the rough spot? (3pt) (c) What is the length of the rough spot? (5pt) (d) 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? (5pt) (e) In this new scenario, to what would the coefficient of friction of the rough patch need to be changed in order for the block to just barely make it through the rough patch? (3pt)

Explanation / Answer

a)

W = (1/2)mv^2 = (0.5) (10) (2.8)^2 = 39.2 J

W = (1/2)kx^2

39.2 = (0.5) (4490) x^2

x = 0.132 m

b)

W_f = (1/2)mv_f^2 - (1/2)mv_i^2

W_f = (0.5)(10)(1.3)^2 - (0.5)(10)(2.8)^2 = 8.45 – 39.2 = -30.75 J

c)

W_f = f_k d

W_f = _k Nd = _k mgd

30.75 = (0.5)(10)(9.8)d

d = 0.628m

d)

Now (1/2)mv'_f^2 - (1/2)mv'_i^2 = - f_k(d/2)

(0.5)(10)(0) - (0.5)(10)v'_i^2 = - (30.75/2)

v'_i^2 = (30.75/2)/(5) = 3.075

(1/2)mv'_i^2 = (1/2)kx'^2

(10)(3.075) = 4490 x'^2

x' = 0.0828 m

e)

(1/2)mv_i^2 = '_k mgd

(0.5)(10)(2.8^2) = '_k (10)(9.8)(0.0828)

'_k = 4.83

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