A motorized winch pulls a 250\text{ kg} equipment sled across a level, frictionless sheet … — Armed Services Vocational Aptitude Battery (ASVAB)

A motorized winch pulls a $250\text{ kg}$ equipment sled across a level, frictionless sheet of ice. If the winch exerts a constant horizontal force of $750\text{ N}$ starting from rest, what is the speed of the sled, in meters per second ( $\text{m/s}$ ), after $4\text{ seconds}$ ?

Answer: 12 m/s

Answer

The sled reaches a speed of

12\text{ m/s}

after

4\text{ seconds}

Applying Newton's second law gives an acceleration of

a = \frac{F}{m} = \frac{750\text{ N}}{250\text{ kg}} = 3\text{ m/s}^2

. Since the sled begins at rest (

v_0 = 0

), its speed after

4\text{ s}

v = at = 3\text{ m/s}^2 \times 4\text{ s} = 12\text{ m/s}

Step-by-Step Solution

Determine the acceleration produced by the net horizontal force

a = \frac{F_{\text{net}}}{m} = \frac{750\text{ N}}{250\text{ kg}} = 3\text{ m/s}^2

According to Newton's second law of motion (

F = ma

), acceleration equals the net force divided by mass.

Calculate final speed from initial rest and constant acceleration over time

v = v_0 + at = 0\text{ m/s} + (3\text{ m/s}^2)(4\text{ s}) = 12\text{ m/s}

Uniform acceleration kinematic equation relates velocity change to acceleration multiplied by elapsed time.

Key Concept

Combining Newton's second law (

F = ma

) with linear kinematics (

v = v_0 + at

) to determine final velocity under constant net force.

Alternative Method

Use the impulse-momentum theorem:

\text{Impulse} = F \cdot \Delta t = \Delta p = m \cdot v

. Substituting known values gives

(750\text{ N})(4\text{ s}) = (250\text{ kg})(v) \implies 3,000\text{ N}\cdot\text{s} = 250\text{ kg} \cdot v \implies v = \frac{3,000}{250} = 12\text{ m/s}

Estimated Time:1m 0s

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Answer

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Common Mistakes

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