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

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 of vi = 3.50 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).

(a) Find the distance of compression d.
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

frictional force acting on the object is given as

f = uk mg = 0.25 x 1.50 x 9.8 = 3.675 N

initial Kinetic energy = (0.5) m Vi2 = (0.5) (1.50) (3.50)2 = 9.1875 J

a)

d = compression of spring

the kinetic energy of the object converts to elastic potential energy of spring and the work done by friction

KE = EPE + W

9.1875 = (0.5) k d2 + f d

9.1875 = (0.5) (50) d2 + 3.675 d

d = 0.54 m

c)

let the speed when in unstretched position be "v"

elastic potential energy converts to kinetic energy and work done by frictional force

EPE = KE + W

(0.5) k d2 = (0.5) m v2 + f d

(0.5) (50) (0.54)2 = (0.5) (1.5) v2 + (3.675) (0.54)'

v = 2.7 m/s

c)

using work-kinetic energy theorem

Work done by friction = change in kinetic energy = final kE - initial KE

-f D = 0 - (0.5) m v2
- 3.675 D = - (0.5) (1.5) (2.7)2

D = 1.5 m

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