91Ó°ÊÓ

The mass of the object is, m=15 kg

The acceleration of the block is,$a=2\mathrm{m}/{\mathrm{s}}^2$

The initial speed is,$v_i=10\mathrm{m}/\mathrm{s}$

The final speed is,$v_f=30\mathrm{m}/\mathrm{s}$

The inclination angle is,$\mathrm{θ}=5°$

We can use the law of conservation to find total energy. Then use the formula for power, which is the ratio of work and time. Instantaneous rate is the multiplication of mechanical energy and velocity.

Formulae:

The total energy of the system,$\mathrm{T}.\mathrm{E}.=\mathrm{K}.\mathrm{E}.+\mathrm{P}.\mathrm{E}.$ (1)

The third equation of kinematic motion,$v_f^2=v_i^2+2ax$ (2)

The average power of a body,$P_{avg}=\frac{w}{t}$ (3)

The instantaneous power of a body,$\mathrm{P}=\mathrm{F}.\mathrm{v}$ (4)

The first equation of kinematic motion,${\mathrm{v}}_{\mathrm{f}}={\mathrm{v}}_{\mathrm{i}}+\mathrm{at}$ (5)

Using equation (2) and the given data, we can get the horizontal distance value as:

$$\begin{gathered} {\left(30\mathrm{m}/\mathrm{s}\right)}^2={\left(10\mathrm{m}/\mathrm{s}\right)}^2+2×2\mathrm{m}/{\mathrm{s}}^2×\mathrm{x} \\ \mathrm{x}=\frac{800{\mathrm{m}}^2/{\mathrm{s}}^2}{4\mathrm{m}/{\mathrm{s}}^2} \\ \mathrm{x}=200\mathrm{m} \end{gathered}$$

Using this value and the value of the angle, we can findtheheight as follows:

$$\begin{gathered} \mathrm{h}=\mathrm{³\ae ²õ¾\pm ²Ôθ} \\ =200\mathrm{m}\sin 5° \\ =17\mathrm{m} \end{gathered}$$

Change in mechanical energy using equations (1) and (2) is given as:

$$\begin{gathered} ∆\mathrm{ME}=0.5{\mathrm{mv}}_{\mathrm{f}}^2-0.5{\mathrm{mv}}_{\mathrm{i}}^2+\mathrm{mgh} \\ =0.5×15\mathrm{kg}\left({\left(30\mathrm{m}/\mathrm{s}\right)}^2-{\left(10\mathrm{m}/\mathrm{s}\right)}^2\right)+15\mathrm{kg}×10\mathrm{m}/{\mathrm{s}}^2×17\mathrm{m} \\ =8550\mathrm{kg}.{\mathrm{m}}^2/{\mathrm{s}}^2\left(\frac{1\mathrm{J}}{1\mathrm{kg}.{\mathrm{m}}^2/{\mathrm{s}}^2}\right) \\ =8550\mathrm{J} \end{gathered}$$

Hence, the value of energy is 8550J.

The time taken for the transfer is given using equation (5):

$$\begin{gathered} 30\mathrm{m}/\mathrm{s}=10\mathrm{m}/\mathrm{s}+\left(2\mathrm{m}/{\mathrm{s}}^2\right) \\ \mathrm{t}=10\mathrm{s} \end{gathered}$$

Now, the average rate of energy transfer can be given using equation (3):

$$\begin{gathered} {\mathrm{P}}_{\mathrm{avg}}=\frac{8550\mathrm{J}}{10\mathrm{s}} \\ =855\mathrm{J}/\mathrm{s}\left(\frac{1\mathrm{W}}{1\mathrm{J}/\mathrm{s}}\right) \\ =855\mathrm{W} \end{gathered}$$

Hence, the required rate of transfer is 855 W.

The instantaneous rate of energy atcan be given using equation (4):

$$\begin{gathered} \mathrm{P}=\left(\mathrm{ma}+\mathrm{mgsin}5\right)×10\mathrm{m}/\mathrm{s} \\ =\left(15\mathrm{kg}×2\mathrm{m}/{\mathrm{s}}^2+15\mathrm{kg}×10\mathrm{m}/{\mathrm{s}}^2\sin 5^{°}\right)×10\mathrm{m}/\mathrm{s} \\ =430\mathrm{W} \end{gathered}$$

Hence, the value of the power is 430W.

The instantaneous rate of energy at 30 m/s can be given using equation (4):

$$\begin{gathered} \mathrm{P}=\left(\mathrm{ma}+\mathrm{mgsin}5\right)×30\mathrm{m}/\mathrm{s} \\ =\left(15\mathrm{kg}×2\mathrm{m}/{\mathrm{s}}^2+15\mathrm{kg}×10\mathrm{m}/{\mathrm{s}}^2\sin 5°\right)×30\mathrm{m}/\mathrm{s} \\ =1300\mathrm{W} \end{gathered}$$

Hence, the value of the power is 1300 W.