Newton's Second Law Physics Example 3

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Example 3

hard
A 5 kg5 \text{ kg} block on a frictionless surface is pulled by a 30 N30 \text{ N} force at an angle of 30°30° above the horizontal. What is the horizontal acceleration?

Solution

  1. 1
    Find the horizontal component of the force: Fx=Fcos⁡θ=30cos⁡30°=30×0.866=25.98 NF_x = F \cos\theta = 30 \cos 30° = 30 \times 0.866 = 25.98 \text{ N}
  2. 2
    Apply Newton's second law horizontally: a=Fxm=25.985=5.2 m/s2a = \frac{F_x}{m} = \frac{25.98}{5} = 5.2 \text{ m/s}^2

Answer

a≈5.2 m/s2a \approx 5.2 \text{ m/s}^2
When a force is applied at an angle, only the horizontal component contributes to horizontal acceleration. The vertical component affects the normal force.

About Newton's Second Law

The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass, with the acceleration pointing.

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More Newton's Second Law Examples

Example 1 easy

A [formula] cart is pushed with a net force of [formula]. What is the acceleration of the cart?

Example 2 medium

A [formula] car accelerates from rest to [formula] in [formula]. What net force is required?

Example 4 medium

A [formula] truck accelerates at [formula] while resistive forces total [formula] opposite the motio

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