<span>1) at rest his
weight is 840 N
=> 840N = mass * g => mass = 840 N / g = 840 N / 9.8 m/s^2 = 85.7 kg
2) as the elevator rises, his weight increases to 1050 N,
The reading of the scale is the norma force of it over the body of the person.
And the equation for the force is: Net force = mass * acceleration = normal force - weight at rest
=> mass * acceleration = 1050 N - 840 N = 210 N
acceleration = 210 N / mass = 210 N / 85.7 kg = 2.45 m/s^2 (upward)
3) when the elevator slows to a stop at the 10th
floor, his weight drops to 588 N
=> mass * acceleration = 588 N - 840 N = - 252 N
=> acceleration = - 252 N / 85.71 kg = - 2.94 m / s^2 (downward)
Answer:
Acceleration at the beginning of the trip 2.45 m/s^2 upward
Acceleration at the end of the trip 2.94 m/s^2 downward
</span>
Answer:
(a) α = -0.16 rad/s²
(b) t = 33.2 s
Explanation:
(a)
Applying 3rd equation of motion on the circular motion of the tire:
2αθ = ωf² - ωi²
where,
α = angular acceleration = ?
ωf = final angular velocity = 0 rad/s (tire finally stops)
ωi = initial angular velocity = 5.45 rad/s
θ = Angular Displacement = (14.4 rev)(2π rad/1 rev) = 28.8π rad
Therefore,
2(α)(28.8π rad) = (0 rad/s)² - (5.45 rad/s)²
α = -(29.7 rad²/s²)/(57.6π rad)
<u>α = -0.16 rad/s²</u>
<u>Negative sign shows deceleration</u>
<u></u>
(b)
Now, we apply 1st equation of motion:
ωf = ωi + αt
0 rad/s = 5.45 rad/s + (-0.16 rad/s²)t
t = (5.45 rad/s)/(0.16 rad/s²)
<u>t = 33.2 s</u>
When angle between them is zero
The size of the forces between you and the planet you're on is
your weight on that planet.
Don't forget that you pull the planet with a force equal to the force
that the planet pulls on you. Your weight on Earth is the same as
the Earth's weight on you !
