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2 years ago

what output force is generated when an input force of 630 n is applied to a machine with a mechanical advantage of 3

Physics

2 answers:

Natali5045456 [20]2 years ago

8 0

Answer : Output force is 1890 N.

Explanation :

Given that,

Input force is 630 N

Mechanical advantage is 3

Mechanical advantage of a machine is defined as the ratio of output force and the input force.

Mathematically it can be written as :

$m=\dfrac{F_o}{F_i}$

Where,

$F_o$ is the output force

$F_i$ is the input force

So, $3=\dfrac{F_o}{630\ N}$

$F_o=1890\ N$

Hence, this is the required solution.

Marta_Voda [28]2 years ago

3 0

The mechanical advantage is the factor by which
the machine multiplies the input force.

If the MA is 3 and the input force is 630N, then
the output force is

(3) x (630N) = 1,890N

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2 years ago

A loaded ore car has a mass of 950 kg. and rolls on rails ofnegligible friction. It starts from rest ans is pulled up a mineshaf

stiks02 [169]

(a) 10241 W

In this situation, the car is moving at constant speed: this means that its acceleration along the direction parallel to the slope is zero, and so the net force along this direction is also zero.

The equation of the forces along the parallel direction is:

$F - mg sin \theta = 0$

where

F is the force applied to pull the car

m = 950 kg is the mass of the car

$g=9.8 m/s^2$ is the acceleration of gravity

$\theta=30.0^{\circ}$ is the angle of the incline

Solving for F,

$F=mg sin \theta = (950)(9.8)(sin 30.0^{\circ})=4655 N$

Now we know that the car is moving at constant velocity of

v = 2.20 m/s

So we can find the power done by the motor during the constant speed phase as

$P=Fv = (4655)(2.20)=10241 W$

(b) 10624 W

The maximum power is provided during the phase of acceleration, because during this phase the force applied is maximum. The acceleration of the car can be found with the equation

$v=u+at$

where

v = 2.20 m/s is the final velocity

a is the acceleration

u = 0 is the initial velocity

t = 12.0 s is the time

Solving for a,

$a=\frac{v-u}{t}=\frac{2.20-0}{12.0}=0.183 m/s^2$

So now the equation of the forces along the direction parallel to the incline is

$F - mg sin \theta = ma$

And solving for F, we find the maximum force applied by the motor:

$F=ma+mgsin \theta =(950)(0.183)+(950)(9.8)(sin 30^{\circ})=4829 N$

The maximum power will be applied when the velocity is maximum, v = 2.20 m/s, and so it is:

$P=Fv=(4829)(2.20)=10624 W$

(c) $5.82\cdot 10^6 J$

Due to the law of conservation of energy, the total energy transferred out of the motor by work must be equal to the gravitational potential energy gained by the car.

The change in potential energy of the car is:

$\Delta U = mg \Delta h$