What speed does it have when it slides back down to its starting point?
- Monday Dec 7,2009 08:00 PM
- By diddy
- In Others
A 2.0 kg wood block is launched up a wooden ramp that is inclined at a 30 degree angle. The block’s initial speed is 10m/s. The coefficient of kinetic friction of wood on wood is .200 . What vertical height does the block reach above its starting point? What speed does it have when it slides back down to its starting point?
10m, Coefficient Of Kinetic Friction, Degree Angle, Initial Speed, Slides, Vertical Height, Wooden Ramp
One Comment
The friction-force is: F = nu*Fg= nu*sin(angle)*m*g= -0.200*2.0*9.81*sin(30).
F = m.a, so a = -0.200*2.0*9.81*sin(30)/2.
Given, s(t) = 1/2*a*t^2 +v0*t. And the speed as a function of time (differentiating s, with respect to t) v(t)=a*t+v0. At the highest point, the speed is zero, so the time will be:
t = -vo/a = -10/(-0.200*2.0*9.81*sin(30)/2) = 10.2 seconds.
The block has slided: s(10.2) = 50.96 meter. But, this is on the ramp. So the vertical height te block reaches above its starting point is 50.96*sin(30)=25,48 meter.
What speed will it have when it reaches it starting point again, for that, we have to find the time when s(t) reaches zero, (use ABC formula). Input that t in our v(t) and voila!
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