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Montano1993 [528]
3 years ago
7

A 1.2 kg block is held at rest against the spring with a force constant k= 730 N/m. Initially, the spring is compressed a distan

ce d. When the block is released, it slides across a surface that is frictionless except for a rough patch of width 5.0 cm that has a coefficient of kinetic friction = 0.44. Find d, such that the block's speed after crossing the rough patch is 2.3 m/s.
Physics
1 answer:
mariarad [96]3 years ago
5 0

Answer:

d = 9.69 cm

Explanation:

given,

mass of the block = 1.2 Kg

spring force constant(k) = 730 N/m

spring is compressed = d = ?

rough patch width = 5 cm

μ_k = 0.44

work done by friction = energy lost

\mu_k mg \times x = \dfrac{1}{2}kd^2-\dfrac{1}{2}mv^2

0.44\times 1.2 \times 9.8 \times 0.05 = \dfrac{1}{2}\times 730 \times d^2-\dfrac{1}{2}\times 1.2 \times 2.3^2

3.432 = 365 d^2

d = \dfrac{3.432}{365}

d = 0.0969 m

d = 9.69 cm

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