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Doss [256]
3 years ago
7

What is the energy put into a machine or device to make it work

Physics
1 answer:
Neporo4naja [7]3 years ago
4 0

Answer:

  • <em>The energy put into a machine or device to make it work is named</em> <u>work input.</u>

Explanation:

<em>Machines </em>are devices to perform tasks with less effort or more easily. They can either increase the force applied or change the direction of the force or permit to apply the force at a distance.

The law of conservation of energy states that the energy is neither created nor destroyed, but when the machines do <em>work</em> some energy is dissipated in the form of heat, light or sound, due, mainly to friction.

In consequence, the <em>work output</em> is less than the <em>work input</em>.

A measure of energy use is efficiency. The efficiency is the ratio of the work output to the work input.

          Efficiency=\dfrac{\text{Work done by the machine}}{\text{Work done on the machine}}

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Please can someone solve this physics question with a good explenation.
zimovet [89]

Answer:

The coefficient of dynamic friction is 0.025.

Explanation:

Given:

Initial speed after the push is 'v' as seen in the graph.

Final speed of the stone is 0 m/s as it comes to rest.

Total distance traveled is, D=29.8\ m

Total time taken is, t_{total}=17.5\ s

Time interval for deceleration is 3.5 to 17.5 s which is for 14 s.

Now, average speed of the stone is given as:

v_{avg}=\frac{D}{t_{total}}=\frac{29.8}{17.5}=1.703\ m/s

Now, we know that, average speed can also be expressed as:

v_{avg}=\frac{v_i+v_f}{2}\\1.703=\frac{v+0}{2}\\v=2\times 1.703=3.41\ m/s

Now, from the graph, the vertical height of the triangles is, v=3.41\ m/s

The deceleration is given as the slope of the line from time 3.5 s to 17.5 s.

Therefore, deceleration is:

a=\frac{\textrm{Vertical height}}{\textrm{Time interval}}\\a=\frac{v-0}{17.5-3.5}\\a=\frac{3.41}{14}=0.244\ m/s^2

Frictional force is the net force acting on the stone. Frictional force is given as:

f=\mu_dN\\Where, \mu_d\rightarrow \textrm{coefficient of dynamic friction}\\N\rightarrow \textrm{Normal force}\\N=mg\\\therefore f=\mu_dmg

Now, from Newton's second law, net force is equal to the product of mass and acceleration.

Therefore,

\mu_dmg=ma\\\mu_d=\frac{a}{g}

Plug in 0.244 for 'a' and 9.8 for 'g'. This gives,

\mu_d=\frac{a}{g}=\frac{0.244}{9.8}=0.025

Therefore, the coefficient of dynamic friction is 0.025.

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