A warehouse forklift applies a constant horizontal forward force of 300\text{ N} to a 120\… — Armed Services Vocational Aptitude Battery (ASVAB)

A warehouse forklift applies a constant horizontal forward force of $300\text{ N}$ to a $120\text{ kg}$ cargo pallet moving across a level floor. A constant opposing friction force of $60\text{ N}$ acts against the pallet. What is the magnitude of the pallet's acceleration in $\text{m/s}^2$ ?

Answer: 2 m/s^2

Answer

The magnitude of the pallet's acceleration is

2\text{ m/s}^2

According to Newton's second law of motion, the acceleration of an object equals the net unbalanced force acting upon it divided by its mass (

a = \frac{F_{\text{net}}}{m}

). The net force is the forward applied force minus opposing friction (

300\text{ N} - 60\text{ N} = 240\text{ N}

). Dividing

240\text{ N}

by the mass of

120\text{ kg}

yields an acceleration of

2\text{ m/s}^2

Step-by-Step Solution

Determine the net horizontal force acting on the object by combining all horizontal force components.

F_{\text{net}} = F_{\text{applied}} - F_{\text{friction}} = 300\text{ N} - 60\text{ N} = 240\text{ N}

Forces in opposite directions subtract to yield the overall resultant force causing acceleration.

Use Newton's second law (

F_{\text{net}} = m \cdot a

) to solve for acceleration (

a

a = \frac{F_{\text{net}}}{m} = \frac{240\text{ N}}{120\text{ kg}} = 2\text{ m/s}^2

Acceleration is directly proportional to net force and inversely proportional to mass.

Key Concept

Newton's Second Law of Motion (

F_{\text{net}} = ma

) with opposing frictional forces

Estimated Time:1m 0s

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