Questions: A block with a mass of 0.440 kg slides, starting from rest, from point A down the frictionless track that includes a loop-the-loop with a radius of r=2.04 m, as shown. The block initially is 5.51 m high from the bottom of the track. What is the minimum speed required for the block to stay on the track at the top of the loop (point B) in m / s ? 4.47 7.75 6.33 9.49 8.95

Solution Steps

To stay on the track at the top of the loop, the block must have a centripetal force equal to or greater than the gravitational force. The minimum speed \( v \) at the top of the loop can be found using the centripetal force equation: \[ \frac{mv^2}{r} = mg \] \[ v^2 = rg \] \[ v = \sqrt{rg} \] Given \( r = 2.04 \, \text{m} \) and \( g = 9.8 \, \text{m/s}^2 \): \[ v = \sqrt{2.04 \times 9.8} \] \[ v = \sqrt{19.992} \] \[ v \approx 4.47 \, \text{m/s} \]

The total mechanical energy at the top of the loop includes both kinetic and potential energy. The height at the top of the loop is \( 2r \): \[ h_{\text{top}} = 2 \times 2.04 = 4.08 \, \text{m} \] The potential energy at the top of the loop is: \[ PE_{\text{top}} = mgh_{\text{top}} = 0.440 \times 9.8 \times 4.08 \] \[ PE_{\text{top}} = 17.6 \, \text{J} \] The kinetic energy at the top of the loop is: \[ KE_{\text{top}} = \frac{1}{2}mv^2 = \frac{1}{2} \times 0.440 \times (4.47)^2 \] \[ KE_{\text{top}} = 4.4 \, \text{J} \] The total mechanical energy at the top of the loop is: \[ E_{\text{top}} = PE_{\text{top}} + KE_{\text{top}} = 17.6 + 4.4 = 22 \, \text{J} \]

The initial potential energy at height \( h \) is: \[ PE_{\text{initial}} = mgh = 0.440 \times 9.8 \times 5.51 \] \[ PE_{\text{initial}} = 23.8 \, \text{J} \] Since the track is frictionless, the initial potential energy is converted into the total mechanical energy at the top of the loop: \[ PE_{\text{initial}} = E_{\text{top}} \] \[ 23.8 \, \text{J} = 22 \, \text{J} \] This confirms that the block has enough energy to reach the top of the loop with the required speed.

Final Answer