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podryga [215]
3 years ago
5

An object of mass m is initially at rest. After a force of magnitude F acts on it for a time T, the object has a speed v. Suppos

e the mass of the object is doubled, and the magnitude of the force acting on it is quadrupled.In terms of T, how long does it take for the object to accelerate from rest to a speed v now?
Physics
1 answer:
adell [148]3 years ago
5 0

Answer:

It takes \frac{1}{2}T to accelerate the object from rest to the speed v.

Explanation:

From Newton's second law:

F=m\cdot a  (1)

and the definition of acceleration,

\displaystyle{a = \frac{\Delta v}{\Delta t}} (2)

we can solve this problem. Putting (2) in (1) we have:

\displaystyle{F = m\cdot \frac{\Delta v}{\Delta t}} and solving for \Delta t and considering the initial time as zero (t_0=0) and the initial velocity also zero (v_0=0) we have:

\displaystyle{T=\frac{mv}{F}}

Now, for a mass m^*= 2m and the F^*=4F we can wrtie the same equation:

\displaystyle{T^*=\frac{m^*v}{F^*}} and substituting m^* and F^*:

\displaystyle{T^*=\frac{2m\cdotv}{4F}=\frac{2}{4}T=\boxed{\frac{1}{2}T}}

So now, it only takes half the time to accelerate the object from rest to the speed v

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\begin{array}{rcl}L & = & \dfrac{\lambda}{2}\\\\\text{1.30 m} & = & \dfrac{\lambda}{2}\\\\\lambda & = & \text{2.60 m}\\\end{array}

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\begin{array}{rcl}v_{3}& = & f_{3}\lambda\\& = & \text{23.40 s}^{-1} \times \text{2.60 m}\\& = & \textbf{60.8 m/s}\\\end{array}

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