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Wednesday, 17 August 2011

Physics GRE - #17

A block of mass $m$ sliding down an incline at constant speed is initially at height $h$ above the ground, as shown in the figure below.

The coefficient of kinetic friction between the mass and the incline is $\mu$. If the mass continues to slide down the incline at a constant speed, how much energy is dissipated by friction by the time the mass reaches the bottom of the incline?

  • $\dfrac{mgh}{\mu}$
  • $mgh$
  • $\dfrac{\mu mgh}{\sin\theta}$
  • $mgh\sin\theta$
  • $0$

Solution :

The answer is $mgh$.

Recall conservation of energy. Since the velocity stays the same at the top and bottom, friction dissipated all of the gravitational potential energy. Thus the value of energy dissipated is $mgh$.
In equation form: \[E_i = E_f \implies mgh+\frac{1}{2}mv_i^2=E_{fr}+\frac{1}{2}mv_f^2.\] But since $v_i=v_f$, \[mgh=E_{fr}.\]


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