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Force P to give block acceleration of 10 m/s²

Short answer

Resolving forces vertically: N = 2500 - 0.5P. Horizontal equation: P cos30° - 0.25N = (2500/9.81)×10. Solving yields 0.991P = 3173.4, so P = 3202 N.

Determine the magnitude of force P required to give the block an acceleration of 10 m/s². Coefficient of friction between the block and the floor is 0.25. [Figure: block of weight W = 2500 N on the floor pulled by force P acting at 30° above the horizontal.]

Reading of the figure

The block (W = 2500 N) rests on a horizontal floor and is pulled by P acting at 30° above the horizontal (the arrow points up and to the right). μ = 0.25, required acceleration a = 10 m/s².

          P  /
            / 30°
   [ W=2500 N ]--->  a = 10 m/s²
 ---------------------------

Equations of motion

  • Vertical (no vertical motion): N + P sin 30° − 2500 = 0, so N = 2500 − 0.5P
  • Horizontal: P cos 30° − μN = m a, with m = 2500/9.81 = 254.84 kg

P(0.8660) − 0.25(2500 − 0.5P) = 254.84 × 10 0.8660P − 625 + 0.125P = 2548.4 0.9910P = 3173.4

P = 3202 N ≈ 3.2 kN

Check: N = 2500 − 1601 = 899 N, friction = 224.8 N, P cos 30° = 2773 N, net = 2548 N = m a ✓.

Answer: P ≈ 3202 N.

✓ Answer checked by Claude Opus 5.5 (high effort)

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