- #1

JMAMA

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- Homework Statement
- A small box with mass 1.20 kg is placed against a light spring that is compressed 0.280 m . The spring, whose other end is attached to a wall, has force constant k = 42.0 N/m . The spring and box are released from rest, and the box travels along a horizontal surface for which the coefficient of kinetic friction with the box is μk = 0.300. When the box has traveled 0.280 m and the spring has reached its equilibrium length, the box loses contact with the spring. What is the maximum speed of the box during its motion?

- Relevant Equations
- E (kinetic) = 1/2mv^2

E (Potential) = 1/2kx^2

F (friction) = u m g

Max speed occurs when all energy has been translated from spring into box.

E (Potential) = 1/2kx^2

E (Potential) = (1/2)(42 N/m)(0.280 m)^2 = 1.6464 N m

Ep = Ek =1/2mv^2

1.6464 N m= 1/2 (1.2 kg) v^2

v = 1.6565 m/s

E (Potential) = 1/2kx^2

E (Potential) = (1/2)(42 N/m)(0.280 m)^2 = 1.6464 N m

Ep = Ek =1/2mv^2

1.6464 N m= 1/2 (1.2 kg) v^2

v = 1.6565 m/s