Questions: A brick of mass 1.00 kg slides down an icy roof inclined at 30.0° with respect to the horizontal. If the brick starts from rest, how fast is it moving when it reaches the edge of the roof 2.87 m away? Ignore friction. Note: Rather than using the kinematic equations described in the beginning of the term, use the law of conservation of energy. m / s

Solution Steps

The brick starts from rest, so its initial kinetic energy is zero. The initial potential energy is given by \( mgh \), where \( h \) is the vertical height of the roof.

The height \( h \) can be found using the distance along the roof and the angle of inclination: \[ h = d \sin \theta \] where \( d = 2.87 \) m and \( \theta = 30.0^\circ \). Thus, \[ h = 2.87 \sin 30.0^\circ = 2.87 \times 0.5 = 1.435 \text{ m} \]

The total mechanical energy is conserved. The initial potential energy is converted into kinetic energy as the brick slides down: \[ mgh = \frac{1}{2} mv^2 \] where \( m = 1.00 \) kg, \( g = 9.81 \) m/s\(^2\), and \( h = 1.435 \) m.

Rearrange the energy conservation equation to solve for \( v \): \[ v = \sqrt{2gh} \] Substitute the known values: \[ v = \sqrt{2 \times 9.81 \times 1.435} = \sqrt{28.1447} \approx 5.305 \text{ m/s} \]

Final Answer