A boy pushes a box of mass 2 kg with a force F⃗ = ( 20î + 10ĵ )N on a frictionless surface. If the box was initially at rest, then ___________ m is displacement along the x-axis after 10s.

Solution

Physics sir · 3 steps

Treat x on its own: only the 20 N part of the force moves the box along x.

1. Split the force. Newton's second law works separately along each axis. Along x, only Fₓ = 20 N acts, and the 10 N along y has no effect on motion along x.
2. Acceleration along x. aₓ = Fₓ/m = 20/2 = 10 m s⁻². The surface is frictionless, so nothing else acts along x.

   x = uₓ t + 1/2 aₓ t² = 0 + 1/2 × 10 × (10)²

3. So. The box starts from rest, so uₓ = 0, and the displacement along the x-axis after 10 s is x = 500 m, which gives 500.

The common slip: 559. Using the magnitude |F⃗| = √500 ≈ 22.4 N instead of Fₓ gives about 559, not 500.

The idea to keep. A force vector is just its components acting at the same time. For motion along one axis, use only that axis's force component in F = ma.

Answer from the NTA final answer key. Solution by PrepWiser.