A motorized winch pulls an 80\text{ kg} cargo pallet horizontally from rest across a concr… — Armed Services Vocational Aptitude Battery (ASVAB)

A motorized winch pulls an $80\text{ kg}$ cargo pallet horizontally from rest across a concrete loading platform. The winch exerts a constant horizontal pulling force of $440\text{ N}$ , while a constant kinetic friction force of $120\text{ N}$ opposes the motion. What is the final speed of the pallet in meters per second ( $\text{m/s}$ ) after being pulled continuously for $5\text{ seconds}$ ?

Answer: 20 m/s

Answer

20\text{ m/s}

The correct speed is

20\text{ m/s}

. By finding the net unbalanced force (

440\text{ N} - 120\text{ N} = 320\text{ N}

), the acceleration of the

80\text{ kg}

pallet is calculated as

a = 320 / 80 = 4\text{ m/s}^2

. Multiplying this constant acceleration by the elapsed time of

5\text{ s}

yields the final speed of

20\text{ m/s}

Step-by-Step Solution

Determine the net horizontal force acting on the cargo pallet by subtracting the opposing friction force from the applied pulling force.

F_{\text{net}} = 440\text{ N} - 120\text{ N} = 320\text{ N}

According to Newton's laws of motion, acceleration is produced only by the unbalanced net force acting on the mass.

Calculate the constant horizontal acceleration using Newton's second law of motion (

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

a = \frac{320\text{ N}}{80\text{ kg}} = 4\text{ m/s}^2

The acceleration of an object is directly proportional to the net force and inversely proportional to its mass.

Compute the final speed after

5\text{ seconds}

using the kinematic velocity-time relationship (

v = v_0 + at

) with an initial speed of

0\text{ m/s}

v = 0\text{ m/s} + (4\text{ m/s}^2)(5\text{ s}) = 20\text{ m/s}

For an object undergoing constant acceleration from rest, final velocity is the product of acceleration and elapsed time.

Key Concept

Newton's Second Law combined with linear kinematics under constant net force

Estimated Time:1m 30s

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