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3 years ago

A ___ force is not path-dependent and does not change the overall mechanical energy of an object.

Physics

2 answers:

finlep [7]3 years ago

6 0

Answer:

conservative

Explanation:

Nonconservative force is the force that depends on a path, however conservative does not depend on a path and it is not associated with the potential energy. When the work is done by an unconservative force, mechanical energy is added or removed. Friction is the best example for a non-conservative force. When these non-conservative forces are acting, the mechanical energy changes but these are not preserved.

hope this helped!

Margaret [11]3 years ago

4 0

Answer: conservative

Explanation:

A conservative force is not path-dependent and does not change the overall mechanical energy of an object.

A conservative force is a force with the property that the total work done in moving a particle between two points is independent of the taken path, e.g. gravitational force or electric force.

A nonconservative force is one for which work depends on the path taken, e.g. frictional force.

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(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:

$\Delta U = mg \Delta h$