Question

A force of \( 20 \text{ N} \) is applied to a \( 5 \text{ kg} \) object at an angle of \( 60^{\circ} \) to the horizontal. What is the horizontal acceleration of the object, ignoring friction?

• A. \( 2 \text{ m/s}^2 \)
• B. \( 4 \text{ m/s}^2 \)
• C. \( 1.73 \text{ m/s}^2 \)
• D. \( 2.5 \text{ m/s}^2 \)

Hint:

For inclined forces, always resolve the force into components first: \[ F_x=F\cos\theta, \qquad F_y=F\sin\theta \] Only the horizontal component contributes to horizontal acceleration.

Answer 1

The Correct Option is A

Concept: When a force is applied at an angle, it is resolved into horizontal and vertical components. The horizontal component is responsible for horizontal acceleration: \[ F_x = F\cos\theta \] Using Newton’s Second Law: \[ a=\frac{F_{\text{net}}}{m} \]

Step 1: Find the horizontal component of the applied force.
Given: \[ F=20\text{ N}, \qquad \theta=60^\circ \] The horizontal component is: \[ F_x=F\cos60^\circ \] Since: \[ \cos60^\circ=\frac{1}{2} \] we get: \[ F_x=20\times \frac{1}{2}=10\text{ N} \]

Step 2: Apply Newton’s Second Law to calculate acceleration.
Mass of the object: \[ m=5\text{ kg} \] Using: \[ a=\frac{F_x}{m} \] Substitute the values: \[ a=\frac{10}{5} \] \[ a=2\text{ m/s}^2 \] Hence, the horizontal acceleration of the object is: \[ \boxed{2\text{ m/s}^2} \]