A 1.50 -kg object slides to the right on a surface having a coefficient of kinet

Question

A 1.50-kg object slides to the right on a surface having a coefficient of kinetic friction 0.250 (Figure a). The object has a speed ofvi = 3.40 m/s when it makes contact with a light spring (Figure b) that has a force constant of 50.0 N/m. The object comes to rest after the spring has been compressed a distance d (Figure c). The object is then forced toward the left by the spring (Figure d) 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 (Figure e).

_______________m

(b) Find the speed v at the unstretched position when the object is moving to the left (Figure d).

_______________m/s

(c) Find the distance D where the object comes to rest.

_______________m

Explanation / Answer

Before touching the spring the object has a Kinetic energy of
KE = 1/2 m*v*v = 1.5 x 3.402/2

                       = 8.67 J

a)

 At the maximum compression 'd', the body comes to rest. By energy conservation all the KE is converted into  the potential energy of the compressed string (kx*x/2)

  So 8.67 = k d2/2 = 25 d2

  So d = 0.589 m = 58.9 cm

b) This is an obvious one if you have your concepts clear.

  Total energy of system = 8.67 J

  In this case spring is unstretched. So Potential Energy = 0

 So 1/2 m V*V = 8.67

 So V = 3.40 m/s

 c) Retarding acceleration -a = f/m = μmg/m = μg = 0.250 x 9.8 = 2.45 m/s/s

   Initial vel V= 3.40m/s

  Finally velocity V' = 0

  Use V'2 - V2 = 2(-a)D

 So D = magnitude of V2/2a = 3.40*3.40/(2*2.45) = 2.36 m.

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