Questions: Block 1 and block 2 have the same mass, m, and are released from the top of two inclined planes of the same height making 30° and 60° angles with the horizontal direction, respectively. If the coefficient of friction is the same in both cases, which of the blocks is going faster when it reaches the bottom of its respective incline? We must know the actual masses of the blocks to answer. Both blocks have the same speed at the bottom. Block 1 is faster. Block 2 is faster.

Solution Steps

Both blocks are subject to gravitational force, normal force, and frictional force. The gravitational force can be decomposed into components parallel and perpendicular to the inclined plane.

For block 1 on the $30^{\circ}$ incline:

• The parallel component of gravitational force: \( mg \sin(30^{\circ}) = \frac{1}{2}mg \)
• The normal force: \( mg \cos(30^{\circ}) = \frac{\sqrt{3}}{2}mg \)
• The frictional force: \( \mu mg \cos(30^{\circ}) = \mu \frac{\sqrt{3}}{2}mg \)

The net force along the incline for block 1: \[ F_{\text{net,1}} = mg \sin(30^{\circ}) - \mu mg \cos(30^{\circ}) = \frac{1}{2}mg - \mu \frac{\sqrt{3}}{2}mg \]

The acceleration \(a_1\) of block 1: \[ a_1 = \frac{F_{\text{net,1}}}{m} = g \left( \frac{1}{2} - \mu \frac{\sqrt{3}}{2} \right) \]

For block 2 on the $60^{\circ}$ incline:

• The parallel component of gravitational force: \( mg \sin(60^{\circ}) = \frac{\sqrt{3}}{2}mg \)
• The normal force: \( mg \cos(60^{\circ}) = \frac{1}{2}mg \)
• The frictional force: \( \mu mg \cos(60^{\circ}) = \mu \frac{1}{2}mg \)

The net force along the incline for block 2: \[ F_{\text{net,2}} = mg \sin(60^{\circ}) - \mu mg \cos(60^{\circ}) = \frac{\sqrt{3}}{2}mg - \mu \frac{1}{2}mg \]

The acceleration \(a_2\) of block 2: \[ a_2 = \frac{F_{\text{net,2}}}{m} = g \left( \frac{\sqrt{3}}{2} - \mu \frac{1}{2} \right) \]

To determine which block is faster at the bottom, we compare \(a_1\) and \(a_2\): \[ a_1 = g \left( \frac{1}{2} - \mu \frac{\sqrt{3}}{2} \right) \] \[ a_2 = g \left( \frac{\sqrt{3}}{2} - \mu \frac{1}{2} \right) \]

Since \(\sin(60^{\circ}) = \frac{\sqrt{3}}{2}\) is greater than \(\sin(30^{\circ}) = \frac{1}{2}\), and assuming \(\mu\) is the same for both blocks, \(a_2\) will generally be greater than \(a_1\).

Final Answer