If an icy surface means no friction, then Newton's second law tells us the net forces on either block are
• <em>m</em> = 1 kg:
∑ <em>F</em> (parallel) = <em>mg</em> sin(45°) - <em>T</em> = <em>ma</em> … … … [1]
∑ <em>F</em> (perpendicular) = <em>n</em> - <em>mg</em> cos(45°) = 0
Notice that we're taking down-the-slope to be positive direction parallel to the surface.
• <em>m</em> = 0.4 kg:
∑ <em>F</em> (vertical) = <em>T</em> - <em>mg</em> = <em>ma</em> … … … [2]
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Adding equations [1] and [2] eliminates <em>T</em>, so that
((1 kg) <em>g</em> sin(45°) - <em>T </em>) + (<em>T</em> - (0.4 kg) <em>g</em>) = (1 kg + 0.4 kg) <em>a</em>
(1 kg) <em>g</em> sin(45°) - (0.4 kg) <em>g</em> = (1.4 kg) <em>a</em>
==> <em>a</em> ≈ 2.15 m/s²
The fact that <em>a</em> is positive indicates that the 1-kg block is moving down the slope. We already found the acceleration is <em>a</em> ≈ 2.15 m/s², which means the net force on the block would be ∑ <em>F</em> = <em>ma</em> ≈ (1 kg) (2.15 m/s²) = 2.15 N directed down the slope.
