A 1.40 kg mass slides to the right on a surface having a coefficient of kinetic

Question

A 1.40 kg mass slides to the right on a surface having a coefficient of kinetic friction 0.250. The object has a speed of vi = 3.50 m/s when it makes contact with a light spring that has a force constant of 50.0 N/m. The object comes to rest after the spring has been compressed a distance d. The object is then forced toward the left by the spring and continues to move in that direction beyond the spring's unstretched position. Finally, the object comes to rest a distance D to the left of the unstretched spring.

Explanation / Answer

A) apply law o conservation of energy

KE = 0.5*k*d^2 + (mu*m*g*d)

KE = 0.5*m*v^2 = 0.5*1.4*3.5^2 =8.575 J

then 8.575 = 0.5*50*d^2 + (0.25*1.4*9.8*d)

25*d^2 + 3.43*d -8.575 = 0

solving we get d = 0.52 m
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apply law of conservation of energy

0.5*k*d^2 = 0.5*m*v^2+(mu*m*g*d)

0.5*50*0.52^2 = 0.5*1.4*v^2 + (0.25*1.4*9.8*0.52)

v = 2.66 m/s
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accelaration = (v^2-u^2)/(2*D) = -2.66^2/(2*D)

But F = m*a

mu*m*g = m*a

mu*g = a = -2.66^2/(2*D)

0.25*9.8 = 2.66^2/(2*D)

D =2.66^2/(2*0.25*9.8) = 1.444m

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