Question

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. ​

Answer 1

#A

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

Use newtons second law

`\\ \rm\longmapsto Force=ma`

`\\ \rm\longmapsto Force=4.4(-1.74)`

`\\ \rm\longmapsto Force=-7.656N`

#B

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

`\\ \rm\longmapsto a=(v-u)/(t)`

`\\ \rm\longmapsto u=v-at`

`\\ \rm\longmapsto u=0-(-1.74)(1.27)`

`\\ \rm\longmapsto u=1.74(1.27)`

`\\ \rm\longmapsto u=2.2m/s`