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A man throws the bricks to a height of 1...

A man throws the bricks to a height of 12 m where they reach with a speed of `12 m//s`. If he throws the bricks such that they just reach that height, what percentage of energy will be saved? `(g = 9.8 m//s^(2))`

A

0.29

B

0.46

C

0.38

D

0.5

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem step by step, we will analyze two cases: 1. When the bricks are thrown to a height of 12 m with a final speed of 12 m/s. 2. When the bricks are thrown to the same height but just reach that height (final speed = 0 m/s). ### Step 1: Calculate the total energy in Case 1 In Case 1, the total energy (E1) can be calculated using the formula: \[ E_1 = \text{Potential Energy} + \text{Kinetic Energy} \] Where: - Potential Energy (PE) = mgh - Kinetic Energy (KE) = \(\frac{1}{2} mv^2\) Given: - \(g = 9.8 \, \text{m/s}^2\) - \(h = 12 \, \text{m}\) - \(v = 12 \, \text{m/s}\) Substituting the values: \[ E_1 = mgh + \frac{1}{2} mv^2 \] \[ E_1 = mg(12) + \frac{1}{2} m(12^2) \] \[ E_1 = m(9.8 \times 12) + \frac{1}{2} m(144) \] \[ E_1 = m(117.6 + 72) = m(189.6) \] ### Step 2: Calculate the total energy in Case 2 In Case 2, where the bricks are thrown to the same height but just reach that height (final speed = 0 m/s): \[ E_2 = mgh + \frac{1}{2} mv^2 \] Since \(v = 0\): \[ E_2 = mgh + 0 \] \[ E_2 = mg(12) \] \[ E_2 = m(9.8 \times 12) = m(117.6) \] ### Step 3: Calculate the energy saved The energy saved (W) can be calculated as: \[ W = E_1 - E_2 \] Substituting the values we found: \[ W = m(189.6) - m(117.6) \] \[ W = m(189.6 - 117.6) = m(72) \] ### Step 4: Calculate the percentage of energy saved The percentage of energy saved can be calculated using the formula: \[ \text{Percentage Saved} = \frac{W}{E_1} \times 100 \] Substituting the values: \[ \text{Percentage Saved} = \frac{m(72)}{m(189.6)} \times 100 \] The mass \(m\) cancels out: \[ \text{Percentage Saved} = \frac{72}{189.6} \times 100 \] Calculating this gives: \[ \text{Percentage Saved} \approx 37.97\% \] ### Final Answer Thus, the percentage of energy saved is approximately **38%**. ---
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