Question

1. 30,000 joules of work are done lifting a 1,500-kg wind turbine blade in the air. what height is the turbine lifted to?

Explanation:

Step1: Recall the work - energy formula for lifting

The work done \( W \) in lifting an object is equal to the change in gravitational potential energy, which is given by the formula \( W=mgh \), where \( m \) is the mass of the object, \( g \) is the acceleration due to gravity (we take \( g = 9.8\space m/s^{2}\) or approximately \( 10\space m/s^{2}\) for simplicity, here we will use \( g=9.8\space m/s^{2}\) first and then check with \( g = 10\space m/s^{2}\) as well), and \( h \) is the height to which the object is lifted. We need to solve for \( h \), so we can rearrange the formula to \( h=\frac{W}{mg} \).

Step2: Identify the given values

We are given that \( W = 30000\space J\) (Joules), \( m=1500\space kg\), and \( g = 9.8\space m/s^{2}\) (acceleration due to gravity).

Step3: Substitute the values into the formula

Substitute \( W = 30000\), \( m = 1500\) and \( g=9.8\) into the formula \( h=\frac{W}{mg}\):
\( h=\frac{30000}{1500\times9.8}\)
First, calculate the denominator: \( 1500\times9.8=14700\)
Then, calculate the numerator divided by the denominator: \( h=\frac{30000}{14700}\approx2.04\space m\)

If we use \( g = 10\space m/s^{2}\) for approximation:
\( h=\frac{30000}{1500\times10}=\frac{30000}{15000} = 2\space m\)

Answer:

If \( g = 9.8\space m/s^{2}\), the height \( h\approx2.04\space m\); if \( g = 10\space m/s^{2}\), the height \( h = 2\space m\) (the answer is often approximated to \( 2\space m\) when using \( g = 10\space m/s^{2}\) for simplicity in some cases).