The filament used in light bulb is Tungsten(W).. It's a metal with melting point of about 3300 °C.

**A 0.293-kg particle undergoes simple harmonic motion along the horizontal x-axis between the points x1 = -0.285 m and x2 = 0.395 m. The period of oscillation is 0.641 s. Find the frequency, f, the equilibrium position, xeq, the amplitude, A, the maximum speed, vmax, the maximum magnitude of acceleration, amax, the force constant, k, and the total mechanical energy, Etot.**

That being said,

f=1/T=1/0.641=1.56 s

xeq=(x1+x2)/2=(-0.285+0.395)/2=0.055 m

A=(x2-x1)/2=(0.395+0.285)/2=0.340 m

x(t)=Acos(wt+ϕ)

v(t)=-wAsen(wt+ϕ)

w=2πf=2*3.14*1.56=9.8 rad/s

vmax=-wA=9.8*0.340=-3.3 m/s

a(t)=-w^2Acos(wt+ϕ)

amax=-w^2A=96.04*0.340=-32.6 m/s^2

w=(k/m)^1/2 => k=mw^2=0.293*96.04=28.1 N/m

Etot=1/2kA^2

Etot=0.5*28.1*0.116=1.63 J

I assume you mean that the car's motor is not running ... the car is just

sitting there.

If that's so, then the car's mechanical energy is just like the mechanical

energy of any other object. It has potential energy if it's in a high place

from which it can roll or fall, and it has kinetic energy if it's moving.

-- If you make the car move by pushing it, then you gave it kinetic energy

that it didn't have while it was just sitting there.

-- If it's already moving slowly, and you're able to make it move faster by

pushing, then you increased its kinetic energy.

-- If you're able to push it up a hill, no matter how small the hill is but just

to any higher place, then you gave it more gravitational potential energy

than it had before you came along.

In all of these cases, if you exert a force and keep exerting it through some

distance while the car moves, then you have done "work", which is just

another name for mechanical energy, and your work adds to the mechanical

energy of the car.

But if you didn't move the car, then no matter how hard you pushed, no work

was done, and the car's mechanical energy didn't change.

Solution:

We have,

Power [P] = 25000 Watt

Mass [m] = 6000 kg

Height [h] = 20 metres

Time [t] = ?

Now,

P = W/t = F x d/t = mxgx h/t

Or, 25000 = 6000 x 10 x 20/25000 [.......g = 10

m/s^2]

Or, t = 6000 x 10 x 20/25000

Or, t = 1200/25

Therefore, t = 48 second

Hence, the required time for the crane to lift the load is 48 seconds.