To find the velocity of the roller coaster car at point 2 using the Law of Conservation of Mechanical Energy, we start by noting that the total mechanical energy is conserved when there are no non-conservative forces like friction.
At point 1 (the top of the track where the height is 25 m), the mechanical energy consists of gravitational potential energy (PE) since the velocity is 0 m/s (kinetic energy (KE) is 0). The gravitational potential energy is given by the equation:
\[ PE = mgh \] Where:
- \( m = 100 , \text{kg} \)
- \( g = 9.81 , \text{m/s}^2 \)
- \( h = 25 , \text{m} \)
Calculating the potential energy at point 1:
\[ PE_1 = 100 , \text{kg} \cdot 9.81 , \text{m/s}^2 \cdot 25 , \text{m} = 24525 , \text{J} \]
At point 2 (at height 10 m), the mechanical energy will consist of both potential energy and kinetic energy.
Calculating the potential energy at point 2:
\[ PE_2 = mgh = 100 , \text{kg} \cdot 9.81 , \text{m/s}^2 \cdot 10 , \text{m} = 9810 , \text{J} \]
Now, using conservation of mechanical energy:
\[ PE_1 = PE_2 + KE_2 \] Where \( KE_2 \) (kinetic energy at point 2) is given by:
\[ KE = \frac{1}{2} mv^2 \]
So we can write:
\[ 24525 , \text{J} = 9810 , \text{J} + \frac{1}{2} \cdot 100 , \text{kg} \cdot v_2^2 \]
Now, rearranging the equation to solve for \( v_2 \):
\[ 24525 , \text{J} - 9810 , \text{J} = \frac{1}{2} \cdot 100 , \text{kg} \cdot v_2^2 \]
Calculating the left side:
\[ 14715 , \text{J} = \frac{100}{2} v_2^2 \] \[ 14715 , \text{J} = 50 v_2^2 \]
Now, divide both sides by 50:
\[ v_2^2 = \frac{14715 , \text{J}}{50} = 294.3 \]
Finally, take the square root to find \( v_2 \):
\[ v_2 = \sqrt{294.3} \approx 17.2 , \text{m/s} \]
Thus, the velocity of the car at point 2 is 17.2 m/s.
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