A mechanical test cart accelerates horizontally from rest to a speed of v, requiring 400\t… — Armed Services Vocational Aptitude Battery (ASVAB)

A mechanical test cart accelerates horizontally from rest to a speed of $v$ , requiring $400\text{ J}$ of net work. How much additional net work must be performed on the cart to further increase its speed from $v$ to $3v$ ?

A
$1,200\text{ J}$
B
$3,600\text{ J}$
$3,200\text{ J}$ Answer
D
$800\text{ J}$

Answer

An additional

3,200\text{ J}

of net work is required.

According to the Work-Energy Theorem, the net work performed on an object equals its change in kinetic energy (

W_{\text{net}} = \Delta \text{KE}

). Accelerating from rest to speed

v

requires

400\text{ J}

, meaning

\frac{1}{2}mv^2 = 400\text{ J}

. At triple the speed (

3v

), the total kinetic energy becomes

\frac{1}{2}m(3v)^2 = 9\left(\frac{1}{2}mv^2\right) = 9 \times 400\text{ J} = 3,600\text{ J}

. The additional work needed to increase speed from

v

3v

is the difference between these two energy states:

3,600\text{ J} - 400\text{ J} = 3,200\text{ J}

Step-by-Step Solution

Express the initial kinetic energy and relate it to initial work done.

\text{KE}_1 = \frac{1}{2}mv^2 = 400\text{ J}

By the Work-Energy Theorem, the net work done on an object starting from rest equals its final kinetic energy.

Calculate the total kinetic energy of the cart when its speed reaches

3v

\text{KE}_2 = \frac{1}{2}m(3v)^2 = 9\left(\frac{1}{2}mv^2\right) = 9 \times 400\text{ J} = 3,600\text{ J}

Kinetic energy depends on the square of velocity, so tripling the velocity increases kinetic energy by a factor of

3^2 = 9

Determine the additional net work needed to transition from speed

v

3v

W_{\text{additional}} = \text{KE}_2 - \text{KE}_1 = 3,600\text{ J} - 400\text{ J} = 3,200\text{ J}

The additional work required equals the change in kinetic energy between the two states.

Key Concept

Work-Energy Theorem and Velocity-Squared Relationship of Kinetic Energy

Estimated Time:1m 15s

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