Recent content by Aristarchus_

Friction, and the normal force from the motorbike, and friction from the motorbike

What is your method of solving this problem?

What answer do you get?

I wrote that the first frictional force when the motorbike is moving on top of the plate is ##R_1 = u*(m+m_p)*g##, where ##u=0.18## times the sum of the masses (m_p -mass of the plate) times the acceleration of gravity. Thus the first work done is R_1*(3m) Then for the second situation, when the...

Fixed it. Thanks

I reason the frictional force on the plate from the ice is doing work first 3 meters (while the motorbike is moving on top) and then an "x" distance after the motorbike has left it. Does anybody have an idea of how one might solve this problem?

d

Nothing other than mass

e

My first attempt was ##... + n^{2} + (n+1)^{2} > \frac {1}{3} n^{3} + (n+1)^{2}## then we must show that ##\frac {1}{3} n^{3} + (n+1)^{2} > \frac {1}{3} (n+1)^{3}## We evaluate both sides and see that the LHS is indeed bigger than RHS. However, this solution is inconsistent so I am asking for...

A maybe better solution...

I took the derivative of ##a_1 \cdot \frac{k^{n}-1}{k-1}##. But how do I determine the minimum of ##S(k) = \dfrac{a_1}{1-k}##? What did you have in mind?

Which expression and what do you get then?

##a_1## cancels out when the derivative is set to 0...

S = a_1 \cdot (k^{n} - 1/ k-1) , this in respect to k and setting n = 1