Question

{
{"An object of mass, m=5, at the origin has a velocity of (12 i - 18 \
j)m/s at t = 0. It is accelerated at a constant rate for 5 seconds \
after which it has a velocity of (5 i - 7 j)m/s."},
{"1. What is the magnitude of the resultant force acting on the \
object during this time interval?
TraditionalForm]\)= .....................N"},
{"2. How far is it from the origin after 3 seconds?
{"3. What is the speed after 3 seconds? Sp = \
.......................m/s"}
}

Answer 1

Answer on Question#52857 - Physics - Atomic Physics

An object of mass, $m = 5\mathrm{kg}$, at the origin has a velocity of $\pmb{v}_i = (12\pmb{i} - 18\pmb{j})\frac{\mathrm{m}}{\mathrm{s}}$ at $t = 0$. It is accelerated at a constant rate for 5 seconds after which it has a velocity of $\pmb{v}_f = (5\pmb{i} - 7\pmb{j})\frac{\mathrm{m}}{\mathrm{s}}$.

1. What is the magnitude of the resultant force acting on the object during this time interval?

2. How far is it from the origin after 3 seconds?

3. What is the speed after 3 seconds?

Solution:

1. Action of the force on the object changes its momentum as follows

$$\int_{t_i}^{t_f} \boldsymbol{F} dt = \Delta \boldsymbol{p},$$

where $t_i, t_f$ – are initial and final times, $\pmb{F}$ – is the resultant force, $\pmb{p} = m\pmb{v}$ – is the momentum of the object. Since the $\pmb{F}$ is constant (object is accelerated at a constant rate), $t_i = 0, t_f = 5\mathrm{s}, \Delta \pmb{p} = m(\pmb{v}_f - \pmb{v}_i)$, we obtain

$$\begin{array}{l} \boldsymbol{F} \left(t_f - t_i\right) = m \left(\boldsymbol{v}_f - \boldsymbol{v}_i\right) \\ F = \frac{m \left(\boldsymbol{v}_f - \boldsymbol{v}_i\right)}{t_f - t_i} = \frac{5\mathrm{kg} \cdot \left(5i - 7j - (12i - 18j)\right) \frac{\mathrm{m}}{\mathrm{s}}}{5\mathrm{s} - 0} = (-7i + 11j)\mathrm{N} \end{array}$$

The magnitude of the force:

$$| \boldsymbol{F} | = \sqrt{(-7)^2 + (11)^2} \mathrm{N} = \sqrt{170} \mathrm{N} = 13 \mathrm{N}$$

2. The displacement is given by

$$\pmb{s}(t) = \pmb{s}_0 + \pmb{v}_i \cdot t + \frac{\pmb{a} \cdot t^2}{2},$$

where $\pmb{s_0} = \pmb{0}$ – is the initial position of the object, $\pmb{a} = \frac{\pmb{F}}{m} = \frac{(-7i + 11j)\mathrm{N}}{5\mathrm{kg}} = (-1.4i + 2.2j)\frac{\mathrm{m}}{\mathrm{s}^2}$ – is the acceleration of the object, $t$ – is the elapsed time.

Since $t = 3\mathrm{s}$, we obtain

$$\pmb{s}(3\mathrm{s}) = (12i - 18j) \frac{\mathrm{m}}{\mathrm{s}} \cdot 3\mathrm{s} + \frac{(-1.4i + 2.2j) \frac{\mathrm{m}}{\mathrm{s}^2} \cdot (3\mathrm{s})^2}{2} = (29.7i + 63.9j)\mathrm{m}$$

The magnitude of $\pmb{s}$ is:

$$| \pmb{s} | = \sqrt{(29.7)^2 + (63.9)^2} \mathrm{m} = \sqrt{4965.3} \mathrm{m} = 70.5 \mathrm{m}$$

3. The dependence of velocity on time is given by

$$\pmb{v}(t) = \pmb{v}_i + \pmb{a} \cdot t,$$

where $t$ – is elapsed time. Since $t = 3\mathrm{s}$, we obtain

$$\pmb{v}(3\mathrm{s}) = (12i - 18j) \frac{\mathrm{m}}{\mathrm{s}} + (-1.4i + 2.2j) \frac{\mathrm{m}}{\mathrm{s}^2} \cdot 3\mathrm{s} = (7.8i - 11.4j) \frac{\mathrm{m}}{\mathrm{s}}$$

The speed is given by the magnitude of this vector

$$v(3\mathrm{s}) = | \pmb{v}(3\mathrm{s}) | = \sqrt{(7.8)^2 + (-11.4)^2} \frac{\mathrm{m}}{\mathrm{s}} = \sqrt{190.8} \frac{\mathrm{m}}{\mathrm{s}} = 13.8 \frac{\mathrm{m}}{\mathrm{s}}$$

Answer:

1. 13N

2. 70.5m

3. $13.8 \frac{\mathrm{m}}{\mathrm{s}}$

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