Questions: The piano slides 2.6 m down a 23° incline and is kept from accelerating by a man who is pushing back on it parallel to the incline (see the figure(Figure 1)). Ignore friction. Previous Answers Correct Part C Determine the work done on the piano by the force of gravity. Express your answer to two significant figures and include the appropriate units. WG = 3700 J Previous Answers Correct Part D Determine the net work done on the piano. Express your answer to two significant figures and include the appropriate units. Wnet Value Units

Solution Steps

A 370 kg piano slides 3.6 m down a 27° incline and is kept from accelerating by a man who is pushing back on it parallel to the incline. We need to determine the work done by the force of gravity.

The force of gravity acting on the piano can be decomposed into two components: one parallel to the incline and one perpendicular to the incline. The parallel component \( F_{\text{gravity, parallel}} \) is given by: \[ F_{\text{gravity, parallel}} = mg \sin(\theta) \] where \( m = 370 \) kg, \( g = 9.8 \) m/s², and \( \theta = 27° \).

The work done by the force of gravity \( W_{\text{gravity}} \) is the product of the parallel component of the gravitational force and the distance \( d \) the piano slides down the incline: \[ W_{\text{gravity}} = F_{\text{gravity, parallel}} \times d \] \[ W_{\text{gravity}} = (370 \times 9.8 \times \sin(27°)) \times 3.6 \]

Final Answer