Find the velocity and displacement of a motor starting from a fixed position after moving 15 s at an acceleration of 2 m s ^ -2.
Please help me...
1 answer:
You might be interested in
- Force=50N
- Acceleration=6.8m/s^2
Answer:
The system will take approximately 0.255 seconds to reach the (new) equilibrium position.
Explanation:
We notice that block-spring system depicts a Simple Harmonic Motion, whose equation of motion is:
(1)
Where:
- Position of the mass as a function of time, measured in meters.
- Amplitude, measured in meters.
- Spring constant, measured in newtons per meter.
- Mass of the block, measured in kilograms.
- Time, measured in seconds.
- Phase, measured in radians.
The spring is now calculated by Hooke's Law, that is:
(2)
Where:
- Gravitational acceleration, measured in meters per square second.
- Deformation of the spring due to gravity, measured in meters.
If we know that ,
and
, then the spring constant is:
If we know that ,
,
,
and
, then (1) is reduced into this form:
(1)
And now we solve for . Given that cosine is a periodic function, we are only interested in the least value of
such that mass reaches equilibrium position. Then:
The system will take approximately 0.255 seconds to reach the (new) equilibrium position.
Answer:
(a) -472.305 J
(b) 1 m
Explanation:
(a)
Change in mechanical energy equals change in kinetic energy
Kinetic energy is given by
Initial kinetic energy is
Since he finally comes to rest, final kinetic energy is zero because the final velocity is zero
Change in kinetic energy is given by final kinetic energy- initial kinetic energy hence
0-472.305 J=-472.305 J
(b)
From fundamental kinematic equation
Where v and u are final and initial velocities respectively, a is acceleration, s is distance
Making s the subject we obtain
but a=\mu g hence