Well, first of all, one who is sufficiently educated to deal with solving
this exercise is also sufficiently well informed to know that a weighing
machine, or "scale", should not be calibrated in units of "kg" ... a unit
of mass, not force. We know that the man's mass doesn't change,
and the spectre of a readout in kg that is oscillating is totally bogus.
If the mass of the man standing on the weighing machine is 60kg, then
on level, dry land on Earth, or on the deck of a ship in calm seas on Earth,
the weighing machine will display his weight as 588 newtons or as
132.3 pounds. That's also the reading as the deck of the ship executes
simple harmonic motion, at the points where the vertical acceleration is zero.
If the deck of the ship is bobbing vertically in simple harmonic motion with
amplitude of M and period of 15 sec, then its vertical position is
y(t) = y₀ + M sin(2π t/15) .
The vertical speed of the deck is y'(t) = M (2π/15) cos(2π t/15)
and its vertical acceleration is y''(t) = - (2πM/15) (2π/15) sin(2π t/15)
= - (4 π² M / 15²) sin(2π t/15)
= - 0.1755 M sin(2π t/15) .
There's the important number ... the 0.1755 M.
That's the peak acceleration.
From here, the problem is a piece-o-cake.
The net vertical force on the intrepid sailor ... the guy standing on the
bathroom scale out on the deck of the ship that's "bobbing" on the
high seas ... is (the force of gravity) + (the force causing him to 'bob'
harmonically with peak acceleration of 0.1755 x amplitude).
At the instant of peak acceleration, the weighing machine thinks that
the load upon it is a mass of 65kg, when in reality it's only 60kg.
The weight of 60kg = 588 newtons.
The weight of 65kg = 637 newtons.
The scale has to push on him with an extra (637 - 588) = 49 newtons
in order to accelerate him faster than gravity.
Now I'm going to wave my hands in the air a bit:
Apparent weight = (apparent mass) x (real acceleration of gravity)
(Apparent mass) = (65/60) = 1.08333 x real mass.
Apparent 'gravity' = 1.08333 x real acceleration of gravity.
The increase ... the 0.08333 ... is the 'extra' acceleration that's due to
the bobbing of the deck.
0.08333 G = 0.1755 M
The 'M' is what we need to find.
Divide each side by 0.1755 : M = (0.08333 / 0.1755) G
'G' = 9.0 m/s²
M = (0.08333 / 0.1755) (9.8) = 4.65 meters .
That result fills me with an overwhelming sense of no-confidence.
But I'm in my office, supposedly working, so I must leave it to others
to analyze my work and point out its many flaws.
In any case, my conscience is clear ... I do feel that I've put in a good
5-points-worth of work on this problem, even if the answer is wrong .
Answer:
Explanation:
Given that,
Mass attached to spring
M = 0.52kg
Force constant K = 8N/m
Amplitude A = 11.6 cm
a. Maximum speed?
Angular velocity is calculated using
w = √k/m
w = √8/0.52
w = √15.385
w = 3.922rad/s
Then, the relation ship between angular velocity and linear velocity is given as
v = - w•A
v = - 3.922 × 11.6
v = - 45.5 cm/s
Then, the maximum velocity is
vmax = |v|= 45.5cm/s
b. Acceleration a?
Acceleration can be determine using the formula
a = -w²• A
a = -3.922² × 11.6
a = -178.46 cm/s²
Magnitude of the acceleration is 178.46cm/s²
c. Speed when the object is at 9.6cm from equilibrium position?
Generally,
The position of the object at equilibrium is
x(t) = A•Cos(wt)
x(t) = 11.6 Cos (3.922t)
Then, when x(t) = 9.6cm
9.6 = 11.6 Cos(3.92t)
Cos(3.922t) = 9.6/11.6
Cos(3.922t) = 0.8276
3.922t = ArcCos(0.8276)
Note: the angle is in radiant
3.922t = 0.596
t = 0.596/3.922
t = 0.152 second
Then, v(t) at that time is
v(t) = x'(t) = -11.6×3.92Sin(3.922t)
v(t) = -45.5Sin(3.922t)
Now, when t =0.152
v(t) = -45.5 Sin(3.922×0.152)
v(t) = -45.5Sin(0.596)
v(t) = -25.5 cm/s
Then, it's magnitude is 25.5cm/s
d. Acceleration at same position
t = 0.152s
a(t) = v'(t) = - 45.5×3.922Cos(3.922t)
a(t) = -178.46Cos(3.92t)
a(t) = -178.46 Cos(3.92×0.152)
a(t) = -178.46 Cos(0.596)
a(t) = -147.68 cm/s²
Magnitude of the acceleration is 147.68 cm/s²
Mass of the object m = 2.9 kg
Force F1 = 28.449 N
F1 = m1 x a => a = F / m => 28.449 / 2.9 => a = 9.81, which is gravitational acceleration.
In the same lab, a = g = 9.81, second object F2 = 48.7N = m2 x a
m2 = F2 / a => 48.7 / 9.81 => m2 = 4.96 kg
Mass of the second object m2 = 4.96 kg
<h3><u>Answer;</u></h3>
Radius = 0.0818 m
Angular velocity = 2.775 × 10^7 rad/sec
<h3><u>Explanation;</u></h3>
The mass of proton m=1.6748 × 10^-27 kg;
Charge of electron e= 1.602 × 10^-19 C;
kinetic energy E= 2.7 MeV
= 2.7 × 10^6 × 1.602 × 10^-19 J;
= 4.32 × 10^-13 Joules
But; K.E =0.5m*v^2,
Hence v=√(2K.E/m)
Velocity = 2.27 × 10^7 m/s
Angular velocity, ω = v/r
Therefore; V = ωr
Hence; V = √(2K.E/m) = ωr
r= √(2E/m)/w = √E*√(2*m)/(eB)
= √E * √(2×1.6748×10^-27)/(1.602×10^-19 ×2.9)
but E = 4.32 × 10^-13 Joules
r = 0.0818 m
Angular speed
Angular velocity, ω = v/r , where r is the radius and v is the velocity
Therefore;
Angular velocity = 2.27 × 10^7 / 0.0818 m
= 2.775 × 10^7 rad /sec
Answer:
a) Ep = 5886[J]; b) v = 14[m/s]; c) W = 5886[J]; d) F = 1763.4[N]
Explanation:
a)
The potential energy can be found using the following expression, we will take the ground level as the reference point where the potential energy is equal to zero.
![E_{p} =m*g*h\\where:\\m = mass = 60[kg]\\g = gravity = 9.81[m/s^2]\\h = elevation = 10 [m]\\E_{p}=60*9.81*10\\E_{p}=5886[J]](https://tex.z-dn.net/?f=E_%7Bp%7D%20%3Dm%2Ag%2Ah%5C%5Cwhere%3A%5C%5Cm%20%3D%20mass%20%3D%2060%5Bkg%5D%5C%5Cg%20%3D%20gravity%20%3D%209.81%5Bm%2Fs%5E2%5D%5C%5Ch%20%3D%20elevation%20%3D%2010%20%5Bm%5D%5C%5CE_%7Bp%7D%3D60%2A9.81%2A10%5C%5CE_%7Bp%7D%3D5886%5BJ%5D)
b)
Since energy is conserved, that is, potential energy is transformed into kinetic energy, the moment the harpsichord touches water, all potential energy is transformed into kinetic energy.
![E_{p} = E_{k} \\5886 =0.5*m*v^{2} \\v = \sqrt{\frac{5886}{0.5*60} }\\v = 14[m/s]](https://tex.z-dn.net/?f=E_%7Bp%7D%20%3D%20E_%7Bk%7D%20%5C%5C5886%20%3D0.5%2Am%2Av%5E%7B2%7D%20%5C%5Cv%20%3D%20%5Csqrt%7B%5Cfrac%7B5886%7D%7B0.5%2A60%7D%20%7D%5C%5Cv%20%3D%2014%5Bm%2Fs%5D)
c)
The work is equal to
W = 5886 [J]
d)
We need to use the following equation and find the deceleration of the diver at the moment when he stops his velocity is zero.
![v_{f} ^{2}= v_{o} ^{2}-2*a*d\\where:\\d = 2.5[m]\\v_{f}=0\\v_{o} =14[m/s]\\Therefore\\a = \frac{14^{2} }{2*2.5} \\a = 39.2[m/s^2]](https://tex.z-dn.net/?f=v_%7Bf%7D%20%5E%7B2%7D%3D%20v_%7Bo%7D%20%5E%7B2%7D-2%2Aa%2Ad%5C%5Cwhere%3A%5C%5Cd%20%3D%202.5%5Bm%5D%5C%5Cv_%7Bf%7D%3D0%5C%5Cv_%7Bo%7D%20%3D14%5Bm%2Fs%5D%5C%5CTherefore%5C%5Ca%20%3D%20%5Cfrac%7B14%5E%7B2%7D%20%7D%7B2%2A2.5%7D%20%5C%5Ca%20%3D%2039.2%5Bm%2Fs%5E2%5D)
By performing a sum of forces equal to the product of mass by acceleration (newton's second law), we can find the force that acts to reduce the speed of the diver to zero.
m*g - F = m*a
F = m*a - m*g
F = (60*39.2) - (60*9.81)
F = 1763.4 [N]
