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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]

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