Question

Repeat Problem 83, but now with the block accelerated up a frictionless plane inclined at $\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{°}$to the horizontal.

Short answer

a) The block’s change in mechanical energy is 8550J

b) The average rate at which energy is transferred to the block is 855W

c) The rate at which energy is transferred at 10m/s is 430 W

d) The rate at which energy is transferred at 30m/s is 1300 W

Step-by-step solution

The given data

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,$\theta =5°$

Understanding the concept of energy

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)

a) Calculation of change in mechanical energy of the block

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

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

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

$$\begin{gathered} \mathrm{h}=\mathrm{xsin\theta } \\ =200\mathrm{m}\sin 5° \\ =17\mathrm{m} \end{gathered}$$

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

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

Hence, the value of energy is 8550J.

b) Calculation of the average rate of energy transfer

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.

c) Calculation of energy transfer at 10 m/s

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

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

Hence, the value of the power is 430W.

d) Calculation of energy transfer at 30 m/s

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

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

Hence, the value of the power is 1300 W.