Work-Energy Theorem Physics Example 2

Follow the full solution, then compare it with the other examples linked below.

Example 2

medium
A 1500ย kg1500 \text{ kg} car traveling at 25ย m/s25 \text{ m/s} brakes with a friction force of 7500ย N7500 \text{ N}. How far does it take to stop?

Solution

  1. 1
    By the work-energy theorem: Wnet=ฮ”KEW_{\text{net}} = \Delta KE.
  2. 2
    Initial KE: 12(1500)(625)=468,750ย J\frac{1}{2}(1500)(625) = 468{,}750 \text{ J}. Final KE: 00.
  3. 3
    Work done by friction (negative, opposing motion): W=โˆ’fโ‹…d=โˆ’7500dW = -f \cdot d = -7500d.
  4. 4
    โˆ’7500d=0โˆ’468,750โ€…โ€ŠโŸนโ€…โ€Šd=468,7507500=62.5ย m-7500d = 0 - 468{,}750 \implies d = \frac{468{,}750}{7500} = 62.5 \text{ m}

Answer

d=62.5ย md = 62.5 \text{ m}
The work-energy theorem shows that the braking distance depends on both the initial kinetic energy and the braking force. Doubling the speed would quadruple the stopping distance.

About Work-Energy Theorem

The net work done on an object by all forces acting on it equals the change in its kinetic energy.

Learn more about Work-Energy Theorem โ†’

More Work-Energy Theorem Examples

Example 1 easy

A [formula] box initially at rest is pushed with a net force of [formula] over [formula]. What is th

Example 3 medium

A [formula] ball moving at [formula] is hit by a bat that does [formula] of work on it. What is the

Example 4 hard

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โ† All Work-Energy Theorem Examples Work-Energy Theorem Concept

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