Question

problem 1: a 0.35 kg ball rolls down a ramp from an initial position of (0,10) m to a final position of (3,1) m. find the work done by gravity. does the path taken matter? note that the gravitational acceleration is (0, -9.8) m/s²

Explanation:

Step1: Calculate the change in height

The initial height \(y_{i}=10\) m and the final height \(y_{f} = 1\) m. The change in height \(\Delta y=y_{f}-y_{i}=1 - 10=-9\) m.

Step2: Use the formula for work done by gravity

The formula for work done by gravity \(W = m g\Delta y\), where \(m = 0.35\) kg and \(g=9.8\) m/s².
Substitute the values: \(W=0.35\times9.8\times(- 9)\)

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The magnitude of work done by gravity is \(|W| = 30.87\space J\).

Since gravity is a conservative force, the work done by gravity depends only on the initial and final positions (i.e., the change in height). So, the path taken does not matter.

Answer:

The work done by gravity is \(30.87\space J\) and the path taken does not matter.