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scoray [572]
3 years ago
10

Two hundred newton·meters of work is put into a machine over a distance of 20 meters. The machine does

Physics
2 answers:
joja [24]3 years ago
4 0
200 newton meters is put over the distance of 20 meters, then over the distance of 10 meters 100 newton-meters must be put (simple fraction over 2).
At the same distance of 10 meters but during lifting you need to put 150 newton-meters (which means 50Nm more than during horizontal movement).

According to that you can say that the machine's advantage is horizontal movement because it does smaller work than during lifting objects. That means it consumes then less energy and then it costs less.
rodikova [14]3 years ago
3 0

Answer:

Mechanical advantage of the machine is 1.5

Explanation:

The formula for mechanical advantage is:

MA = OutputForce/InputForce

To calculate this, we need the force of input and the force of output.

The formula to calculate the force given the work (W) and the distance (d) is:

F = W/d

Calculating the input force Fi:

Fi = 200Nm/20m = 10 N

Calculating the output force Fo:

Fo = 150Nm/10m = 15 N

Thus, the mechanical advantage MA is:

MA = 15N / 10N = 1.5

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A 17.6-kg block rests on a horizontal table and is attached to one end of a massless, horizontal spring. By pulling horizontally
leonid [27]

In order get the block up to a speed of 3.58 m/s in 1.77 s, it must undergo an acceleration <em>a</em> of

<em>a</em> = (3.58 m/s) / (1.77 s) ≈ 2.02 m/s²

When the spring is getting pulled, Newton's second law tells us

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∑ <em>F</em> = <em>n</em> - <em>mg</em> = 0

where ∑ <em>F</em> is the net force, <em>n</em> is the magnitude of the normal force, and <em>mg</em> is the weight of the block - it follows that <em>n</em> = <em>mg</em> ; and

• the net horizontal force is

∑ <em>F</em> = <em>F</em> - <em>f</em> = <em>ma</em>

where <em>F</em> is the applied force, <em>f</em> is kinetic friction, <em>m</em> is the block's mass, and <em>a</em> is the acceleration found earlier. <em>F</em> stretches the spring by <em>x</em> = 0.250 m, so we have

<em>F</em> - <em>f</em> = <em>kx</em> - <em>µn</em> = <em>kx</em> - <em>µmg</em> = <em>ma</em>

where <em>k</em> is the spring constant and <em>µ</em> is the coefficient of kinetic friction.

When the block is being pulled at a constant speed, Newton's second law says

• the net vertical force is still

∑ <em>F</em> = <em>n</em> - <em>mg</em> = 0

so that <em>n</em> = <em>mg</em> again; and

• the net horizontal force is

∑ <em>F</em> = <em>F</em> - <em>f</em> = 0

This time, <em>F</em> stretches the spring by <em>y</em> = 0.0544 m, so we have

<em>F</em> - <em>f</em> = <em>ky</em> - <em>µmg</em> = 0

Solve the equations in boldface for <em>k</em> and <em>µ</em> :

<em>kx</em> - <em>µmg</em> = <em>ma</em>

<em>ky</em> - <em>µmg</em> = 0

==>   <em>k</em> (<em>x</em> - <em>y</em>) = <em>ma</em>

==>   <em>k</em> = <em>ma</em> / (<em>x</em> - <em>y</em>)

==>   <em>k</em> = (17.6 kg) (2.02 m/s²) / (0.250 m - 0.0544 m) ≈ 182 N/m

Then

<em>ky</em> - <em>µmg</em> = 0

==>   <em>µ</em> = <em>ky </em>/ (<em>mg</em>)

==>   <em>µ</em> = (182 N/m) (0.0544 m) / ((17.6 kg) <em>g</em>) ≈ 0.0574

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Yuri [45]

Answer:

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...

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