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mrs_skeptik [129]
3 years ago
8

A physicist does 100 joules of work on a simple machine that raise a box of bioks through a heigjt of 0.2metres. if the efficien

cy of the machine is 60%, how much work is converted to thermal energy?​
Physics
1 answer:
11Alexandr11 [23.1K]3 years ago
5 0

Answer:

the work converted to thermal energy is 40 J.

Explanation:

Given;

work done by the physicist,w = 100 J

height through which the book is raised, h = 0.2 m

efficiency of machine = 60% = 0.6

The useful work done by the machine is calculated as;

useful work = 0.6 x 100 = 60 J

The wasted energy = 100 J - 60 J

The wasted energy = 40 J

The wasted energy by the machine is possibly converted to thermal energy by the frictional part of the machine.

Therefore, the work converted to thermal energy is 40 J.

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Answer:

μsmín = 0.1

Explanation:

  • There are three external forces acting on the riders, two in the vertical direction that oppose each other, the force due to gravity (which we call weight) and the friction force.
  • This friction force has a maximum value, that can be written as follows:

       F_{frmax} = \mu_{s} *F_{n} (1)

       where  μs is the coefficient of static friction, and Fn is the normal force,

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  • This force is the only force acting in the horizontal direction, but, at the same time, is the force that keeps the riders rotating, which is the centripetal force.
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       F_{c} =  m* \omega^{2} * r (2)

       where ω is the angular velocity of the riders, and r the distance to the

      center of rotation (the  radius of the circle), and m the mass of the

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     F_{frmax} = m* \mu_{s} * \omega^{2} * r (3)

  • When the riders are on the verge of sliding down, this force must be equal to the weight Fg, so we can write the following equation:

       m* g = m* \mu_{smin} * \omega^{2} * r (4)

  • (The coefficient of static friction is the minimum possible, due to any value less than it would cause the riders to slide down)
  • Cancelling the masses on both sides of (4), we get:

       g = \mu_{smin} * \omega^{2} * r (5)

  • Prior to solve (5) we need to convert ω from rev/min to rad/sec, as follows:

      60 rev/min * \frac{2*\pi rad}{1 rev} *\frac{1min}{60 sec} =6.28 rad/sec (6)

  • Replacing by the givens in (5), we can solve for μsmín, as follows:

       \mu_{smin} = \frac{g}{\omega^{2} *r}  = \frac{9.8m/s2}{(6.28rad/sec)^{2} *2.5 m} =0.1 (7)

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