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. Construct an FBD of the situation at the point when the marble reaches level ground.

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

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

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Аноним Аноним · Answer 1 · 2021-09-21T06:10:31+02:00

#A

Refer to attachment

#B

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]

#C

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]