A block of mass 0.74 kg is suspended by a string which is wrapped so that it is

Question

A block of mass 0.74 kg is suspended by a string which is wrapped so that it is at a radius of 0.068 m from the center of a pulley. The moment of inertia of the pulley is 5.80

A block of mass 0.74 kg is suspended by a string which is wrapped so that it is at a radius of 0.068 m from the center of a pulley. The moment of inertia of the pulley is 5.80?10-3kg·m2. There is friction as the pulley turns. The block starts from rest, and its speed after it has traveled downwards a distance of D= 0.83 m, is 1.888 m/s. Calculate the amount of energy dissipated up to that point.

Explanation / Answer

  Use work energy to solve this. Let U be potential energy = m*g*h and K be kinetic energy = 1/2*m*v^2 or 1/2*I*?^2 (Note without slipping v = r*?) and W (friction) is the work by friction which is the energy dissipated

Let the initial U of the block = 0

so

(K + U) initial + W(friction) = (K + U) final

K initial = 0 and U initial = 0

therefore W = 1/2*m*v^2 + 1/2*I*? + m*g*h

W = 1/2*0.74kg*1.888^2 + 1/2*0.0058*(1.888/0.068)^2 + 0.74*9.8*(-0.83)

= 1.3188 + 2.23554 - 6.0253 = -2.471J

So there are 2.471J of energy dissipated

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