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siniylev [52]
3 years ago
5

If a single constant force acts on an object that moves on a straight line, the object's velocity is a linear function of time.

The equation v = v_i + at gives its velocity v as a function of time, where a is its constant acceleration. What if velocity is instead a linear function of position? Assume that as a particular object moves through a resistive medium, its speed decreases as described by the equation v = v_i - kx, where k is a constant coefficient and x is the position of the object. Find the law describing the total force acting on this object. (Enter an expression for the magnitude of the total force. Use the following as necessary: m, k, and v.)
Physics
1 answer:
olya-2409 [2.1K]3 years ago
5 0

Answer:

F=mkv

Explanation:

Given that

v = v_i - kx

We know that acceleration a given as

a=\dfrac{dv}{dt}

v = v_i - kx

\dfrac{dv}{dt}=\dfrac{dv_i}{dt}-k\dfrac{dx}{dt}

\dfrac{dv}{dt}=0-k\dfrac{dx}{dt}

We know that

F=m\dfrac{dv}{dt}

F=-mk\dfrac{dx}{dt}

F=-mkv

So the magnitude of force F

F=mkv

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Explanation:

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A hotel elevator ascends 200m with maximum speed of 5m/s. Its acceleration and deceleration both have a magnitude of 1.0m/s2. Pa
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Answer:

45 s .

Explanation:

The accelerator will first accelerate , then move with uniform velocity and at last it will decelerate to rest .

displacement s = ?

acceleration a = 1 m /s²

Final speed v = 5 m/s

initial speed u = 0

v² = u² + 2as

5² = 0 + 2 x 1 x s

s = 12.5  m

B)  Let time of acceleration or deceleration be t

v = u + a t

5 = 0 + 1 t

t = 5 s

Similarly displacement during deceleration = 12.5 m

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A spaceship negotiates a circular turn of radius 2925 km at a speed of 29960 km/h. (a) What is the magnitude of the angular spee
emmainna [20.7K]

a) 0.0028 rad/s

b) 23.68 m/s^2

c) 0 m/s^2

Explanation:

a)

When an object is in circular motion, the angular speed of the object is the rate of change of its angular position. In formula, it is given by

\omega = \frac{\theta}{t}

where

\theta is the angular displacement

t is the time interval

The angular speed of an object in circular motion can also be written as

\omega = \frac{v}{r} (1)

where

v is the linear speed of the object

r is the radius of the orbit

For the spaceship in this problem we have:

v=29,960 km/h is the linear speed, converted into m/s,

v=8322 m/s

r=2925 km = 2.925\cdot 10^6 m is the radius of the orbit

Subsituting into eq(1), we find the angular speed of the spaceship:

\omega=\frac{8322}{2.925\cdot 10^6}=0.0028 rad/s

b)

When an object is in circular motion, its direction is constantly changing, therefore the object is accelerating; in particular, there is a component of the acceleration acting towards the  centre of the orbit: this is called centripetal acceleration, or radial acceleration.

The magnitude of the radial acceleration is given by

a_r=\omega^2 r

where

\omega is the angular speed

r is the radius of the orbit

For the spaceship in the problem, we have

\omega=0.0028 rad/s is the angular speed

r=2925 km = 2.925\cdot 10^6 m is the radius of the orbit

Substittuing into the equation above, we find the radial acceleration:

a_r=(0.0028)^2(2.925\cdot 10^6)=23.68 m/s^2

c)

When an object is in circular motion, it can also have a component of the acceleration in the direction tangential to its motion: this component is called tangential acceleration.

The tangential acceleration is given by

a_t=\frac{\Delta v}{\Delta t}

where

\Delta v is the change in the linear speed

\Delta  t is the time interval

In this problem, the spaceship is moving with constant linear speed equal to

v=8322 m/s

Therefore, its linear speed is not changing, so the change in linear speed is zero:

\Delta v=0

And therefore, the tangential acceleration is zero as well:

a_t=\frac{0}{\Delta t}=0 m/s^2

5 0
3 years ago
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