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Physics 19 Online
OpenStudy (pah001):

A 15 kg curling rock is released with a speed of 2.02 m/s and slides to a stop in 11 meters. What force of friction acts on the rock?

OpenStudy (diglet):

Ok,I'll give it a shot. Firstly we need to calculate time taken to stop. |dw:1477447214237:dw|

OpenStudy (diglet):

The we use this time to calculate the acceleration using v = u + at. a = (v-u) / t a = (0 - 2.02) / 10.89 a = -0.185 ms^-2

OpenStudy (diglet):

The force that causes and acceleration is given by the formula f=ma, f is the friction force in this case. f = 15kg / (-) 0.18549... f= (-)80.9 N This force is positive or negative depending on which way you look at it. Hope this is correct and that it helps

OpenStudy (irishboy123):

|dw:1477518297765:dw|

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