The kinetic energy \( K \) of a moving object varies directly with the square of its speed \( v \). If \( K=450 \mathrm{~J} \) when \( v=30 \mathrm{~m} / \mathrm{s} \), what is the speed when \( K=800 \mathrm{~J} \) ? \( 35.5 \mathrm{~m} / \mathrm{s} \) \( 37.5 \mathrm{~m} / \mathrm{s} \) \( 42.5 \mathrm{~m} / \mathrm{s} \) \( 40 \mathrm{~m} / \mathrm{s} \)

To solve for the speed when \( K = 800 \, \text{J} \), we first need to establish the relationship between kinetic energy and speed. Kinetic energy is given by the equation \( K = \frac{1}{2} mv^2 \). Since \( K \) varies directly with \( v^2 \), we can write the proportionality as \( K = k v^2 \) where \( k \) is a constant. From the given values, we know: \[ 450 = k (30^2) \quad \Rightarrow \quad 450 = k (900) \quad \Rightarrow \quad k = \frac{450}{900} = 0.5. \] Now we can use this constant to find the speed when \( K = 800 \, \text{J} \): \[ 800 = 0.5 v^2 \quad \Rightarrow \quad 800 = \frac{1}{2} v^2 \quad \Rightarrow \quad v^2 = 1600 \quad \Rightarrow \quad v = \sqrt{1600} = 40 \, \text{m/s}. \] Thus, the speed when \( K = 800 \, \text{J} \) is \( 40 \, \text{m/s} \). In the realm of physics, the direct relationship between kinetic energy and the square of velocity highlights a curious fact: a small increase in speed can lead to a significant increase in kinetic energy! For example, if your speed doubles, your kinetic energy quadruples. It's a thrilling reminder of the power harnessed in motion! When you're tackling physics problems like this, take a moment to jot down the known values and formulate your equations clearly. A common mistake is skipping steps, which can lead to confusion. Also, keep track of your units; ensuring you're consistent will save you from unnecessary errors. Remember, every small calculation matters in this exciting journey through the world of physics!