A hockey puck is hit on a frozen lake and starts moving with aspeed of 12.9 m/s.

Question

A hockey puck is hit on a frozen lake and starts moving with aspeed of 12.9 m/s. Fiveseconds later, its speed is6.60 m/s. (a) Whatis its average acceleration?
m/s2

(b) What is the average value of the coefficient of kineticfriction between puck and ice?


(c) How far does the puck travel during the 5.00 s interval?
m (a) Whatis its average acceleration?
m/s2

(b) What is the average value of the coefficient of kineticfriction between puck and ice?


(c) How far does the puck travel during the 5.00 s interval?
m

Explanation / Answer

   Given      initialvelocity ofpuck   u   =   12.9,   finalvelocity   v   =   6.60 m/s,   time   t   =   5.00s    a.   First equation of motionis          v   =   u    +    a* t          6.60   =   12.9   +   a* 5.00          acceleration   a   =   (6.60- 12.9) / 5.00   =   -1.26   m/s2    - ve sign represents retarded motion.    b.   deacceleration   a   =   * g    =>   coefficient ofkineticfriction      =   a/ g   =   1.26 /9.8   =   0.128    c.   Second equation ofmotion is       distancetravelled   s   =   u* t   +   (1/2) * a *t2    s   =   12.9 *5.0   +   0.5 * (-1.26) *5.02    s   =   48.75   m    s   =   48.75   m

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