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A 4.4 kg marble (really big heavy marble) is accelerating down an incline. When it reaches level ground it slows down to a stop in 1.27 seconds due to friction.


A. If the deceleration of the marble on level ground was 1.74 m/s/s, how much frictional force was present?


B. Calculate the velocity of the marble when it initially reached level ground. ​

Respuesta :

#A

  • Mass=4.4kg
  • acceleration=-1.74m/s^2

Use newtons second law

[tex]\\ \rm\longmapsto Force=ma[/tex]

[tex]\\ \rm\longmapsto Force=4.4(-1.74)[/tex]

[tex]\\ \rm\longmapsto Force=-7.656N[/tex]

#B

  • initial velocity=u
  • Final velocity=v=0
  • Acceleration=a=-1.74m/s^2
  • Time=t=1.27s

[tex]\\ \rm\longmapsto a=\dfrac{v-u}{t}[/tex]

[tex]\\ \rm\longmapsto u=v-at[/tex]

[tex]\\ \rm\longmapsto u=0-(-1.74)(1.27)[/tex]

[tex]\\ \rm\longmapsto u=1.74(1.27)[/tex]

[tex]\\ \rm\longmapsto u=2.2m/s[/tex]

estelamariediocampo3

ANSWER:

#A

#AMass=4.4kg

#AMass=4.4kgacceleration=-1.74m/s^2

#AMass=4.4kgacceleration=-1.74m/s^2Use newtons second law

⟼Force=ma

⟼Force=4.4(−1.74)

⟼Force=−7.656N

⟼Force=−7.656N

#B

#Binitial velocity=u

#Binitial velocity=uFinal velocity=v=0

#Binitial velocity=uFinal velocity=v=0Acceleration=a=-1.74m/s^2

#Binitial velocity=uFinal velocity=v=0Acceleration=a=-1.74m/s^2Time=t=1.27s

⟼a=

⟼a= t

⟼a= tv−u

⟼a= tv−u

⟼u=v−at

⟼u=0−(−1.74)(1.27)

⟼u=0−(−1.74)(1.27)

⟼u=1.74(1.27)

⟼u=1.74(1.27)

⟼u=2.2m/s

⟼u=2.2m/s

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