A mass m = 16 kg rests on a frictionless table and accelerated by a spring with

Question

A mass m = 16 kg rests on a frictionless table and accelerated by a spring with spring constant k = 5052 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 = 4 m/s.

1) How far was the spring stretched from its unstreched length?

2) The mass is measured to leave the rough spot with a final speed vf = 2.3 m/s. How much work is done by friction as the mass crosses the rough spot?

3) What is the length of the rough spot?

4) 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?

Explanation / Answer

GIVEN:

m = 16 kg,k = 5052 N/m,k = 0.52.,v = 4 m/s.

1) How far was the spring stretched from its unstreched length?

work done :W = (1/2)mv^2 = (0.5) *(16)* ((4)^2) = 128 J

W = (1/2)kx^2
128 = (0.5) (5052) x^2
x = 0.225m

2) The mass is measured to leave the rough spot with a final speed vf = 2.3 m/s. How much work is done by friction as the mass crosses the rough spot?

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

W_f = (0.5)(16)(4)^2 - (0.5)(16)(2.3)^2 = 85.68 J

3) What is the length of the rough spot?

W_f = f_k d
W_f = _k Nd = _k mgd
85.68 = (0.52)(16)(9.81)d
d = 1.049 m

4) 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?

Now (1/2)mv'_f^2 - (1/2)mv'_i^2 = - f_k(d/2)
(0.5)(16)(0) - (0.5)(16)v'_i^2 = - (85.68/2)
v'_i = 2.31m/s
(1/2)mv'_i^2 = (1/2)kx'^2
(16)(5.355) = 5052 x'^2
x' = 0.13 m

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