Answer:
W = 172.5 J
Explanation:
given,
mass of the fruit crate = 14.5 kg
initial velocity to lift = 0.500 m/s
increase in the tension = 150 N
lift of crate = 1.15 m
work done by the tension = ?
work done = force x displacement
W = F s cos θ
θ = 0°
W = F s x cos 0
W = 150 x 1.15 x 1
W = 172.5 J
Work done on the crate by the tension force = W = 172.5 J
Answer:
t=37 mins -> 2220sec
We want "T" which is the pendulum time constant
Using this equation
.5A=Ae^(-t/T)
The .5A is half the amplitude
Take ln of both sides to get ride of Ae
=ln(.5)=-2220/T
Now rearrange to = T
T=-2220/ln(.5) = 3202.78sec / 60 secs = 53.38 mins -> first part of the answer.
The second part is really easy. It took 37 mins to decay half way. meaning to decay another half of 50% which equals 25% it will take an additional 37 mins!
Answer:d
Explanation:
Spring is compressed to a distance of x from its equilibrium position
Work done by block on the spring is equal to change in elastic potential energy
i.e. Work done by block 
therefore spring will also done an equal opposite amount of work on the block in the absence of external force
Thus work done by spring on the block 
Thus option d is correct
Answer:
U = 1 / r²
Explanation:
In this exercise they do not ask for potential energy giving the expression of force, since these two quantities are related
F = - dU / dr
this derivative is a gradient, that is, a directional derivative, so we must have
dU = - F. dr
the esxresion for strength is
F = B / r³
let's replace
∫ dU = - ∫ B / r³ dr
in this case the force and the displacement are parallel, therefore the scalar product is reduced to the algebraic product
let's evaluate the integrals
U - Uo = -B (- / 2r² + 1 / 2r₀²)
To complete the calculation we must fix the energy at a point, in general the most common choice is to make the potential energy zero (Uo = 0) for when the distance is infinite (r = ∞)
U = B / 2r²
we substitute the value of B = 2
U = 1 / r²
