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adelina 88 [10]
3 years ago
5

In which one of the following circumstances could mechanical energy not possibly be conserved, even if friction and air resistan

ce are absent?
A car moves down a hill, its velocity continually increasing alongthe way.
A car moves up a hill at a constant velocity.
A car moves along level ground at a constant velocity.
A car moves up a hill, its velocity continually decreasing alongthe way.
Physics
1 answer:
qwelly [4]3 years ago
5 0

Answer:

A car moves up a hill at a constant velocity

Explanation:

Since the velocity is constant, the speed is also constant and so is the kinetic energy. However, total mechanical energy is sum of gravitational potential energy and kinetic energy, and the car is moving up the hill so its potential energy rises.

Thus, in the circumstances described the mechanical energy cannot be conserved.

The correct answer is A car moving up the hill with constant velocity.

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(1 point) A mass m=4kg is attached to both a spring with spring constant k=577N/m and a dash-pot with damping constant c=4N⋅s/m
sasho [114]

Answer:

The function is x = e^(-t/2) * (0.792*sin12t + 5cos12t)

Explanation:

we have to:

m = mass = 4 kg

k = spring constant = 577 N/m

c = damping constant = 4 N*s/m

The differential equation of motion is equal to:

m(d^2x/dt^2) + c(dx/dt) + k*x = 0

Replacing values:

4(d^2x/dt^2) + 4(dx/dt) + 577*x = 0

Thus, we have:

4*x^2 + 4*x + 577 = 0

we will use the quadratic equation to solve the expression:

x = (-4 ± (4^2 - (4*4*577))^1/2)/(2*4) = (-4 ± (-9216))/8 = (1/2)  ± 12i

The solution is equal to:

x = e^(1/2) * (c1*sin12t + c2*cos12t)

x´ = (-1/2)*e^(1/2) * (c1*sin12t + c2*cos12t) + e^(-t/2) * (12*c1*cos12t - 12*c2*sin12t)

We have the follow:

x(0) = 5

e^0(0*c1 + c2) = 5

c2 = 5

x´(0) = 7

(-1/2)*e^0 * (0*c1 + c2) + e^0 * (12*c1 - 0*c2) = 7

(-1/2)*(5) + 12*c1 = 7

Clearing c1:

c1 = 0.792

The function is equal to:

x = e^(-t/2) * (0.792*sin12t + 5cos12t)

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