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anonymous
 one year ago
Two blocks are connected by a string that goes around a pulley. Block 1 is on a slope of angle "theta", and Block 2 is dangling off the edge below the pulley. The slope has a kinetic friction coefficient "mu". Find the change in kinetic energy of Block 1 as it moves a distance "d" up the slope.
anonymous
 one year ago
Two blocks are connected by a string that goes around a pulley. Block 1 is on a slope of angle "theta", and Block 2 is dangling off the edge below the pulley. The slope has a kinetic friction coefficient "mu". Find the change in kinetic energy of Block 1 as it moves a distance "d" up the slope.

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1434668108317:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I have been able to solve this with \[\Delta KE = W_{f}  \Delta PE\], but I don't understand why tension is excluded from the calculations. Could someone help me understand how to approach a problem like this? I also tried calculating the force of the system (m1 and m2 together), and then multiplying that by d, but that didn't work out. Why did that not work?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Hi! dw:1434673440036:dw This is a good question. If we look at this problem using the WorkEnergy theorem: \[\Delta E_K = W_{net}\] the variation of kinetic energy equals the net work (the sum of all works), we need to consider all the forces that are doing work, the forces involved for mass 1 are:  tension \(T_1\)  weight \(m_1g\)  friction \(f\)  Normal \(N\) for mass 2 we have:  tension \(T_2\)  weight \(m_2g\) we know that \(N\) is orthogonal to the displacement of mass 1, so its work is 0. the same happens for one component of the weight of mass 1, so for that wight we only account for \(W_{m_1g} = m_1g\sin{(\theta)}\times d\) the work done by component that is parallel to the plane. the works for tension 1 and friction are: \(W_{T_1}=T_1\times d\) and \(W_{f}=f\times d\) for the second mass we have the work of the weight and tension 2: \(W_{T_2}=T_2\times h\) and \(W_{m_2g}=m_2g\times h\) now we take alook at some conditions, since both blocks are connected by an ideal string we can say that: \(T_1=T_2\) and \(h=d\) so if we add all of the works togheter, \(W_{T_1}\) and \(W_{T_2}\) will cancel out, and we are left only with \(W_{f}\), \(W_{m_1g}\) and \(W_{m_2g}\), the last two are actually equal to the variation of gravitational potential energy \(\Delta E_{P}\), so we have: \[\Delta E_K = W_{f} + \Delta E_{P}\] the sign of the friction term will always be negative. For the potential energy it will depend on the relation between the masses. Hope this answers your question!
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