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  1. Problems
  2. /A 1.5 kg cart moves along the x axis at 2.0 m/s when it passes x = 1 m…
HardPhysicsHigh School (11–12)Mechanics › Work, Energy and PowerNumericalJEE MainNEETBITSATWBJEEBoard ExamAP

A 1.5 kg cart moves along the x axis at 2.0 m/s when it passes x = 1 m…

StemPy LabContributed by StemPy Lab

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A 1.5 kg1.5\ \text{kg}1.5 kg cart moves along the xxx axis at 2.0 m/s2.0\ \text{m/s}2.0 m/s when it passes x=1 mx = 1\ \text{m}x=1 m. From there a force F=2x2+3F = 2x^2 + 3F=2x2+3 newtons (with xxx in metres) pushes it along the axis.

  1. How much work does the force do between x=1 mx = 1\ \text{m}x=1 m and x=4 mx = 4\ \text{m}x=4 m?
  2. How fast is the cart moving at x=4 mx = 4\ \text{m}x=4 m? Ignore friction.

Check your answer

Work it out on paper, then type your number. The answer itself stays hidden.

J
m/s
Need a reminder? The key ideaShowHide

The work of a force that changes along the path is the area under its force-position graph, W=∫F dxW = \int F\,dxW=∫Fdx. For a spring, W=12k(x22−x12)W = \tfrac12 k(x_2^2 - x_1^2)W=21​k(x22​−x12​).

Kinetic energy is K=12mv2K = \tfrac12 mv^2K=21​mv2. The work-energy theorem: the net work done on a body equals its change in kinetic energy, Wnet=12mv2−12mu2W_{\text{net}} = \tfrac12 mv^2 - \tfrac12 mu^2Wnet​=21​mv2−21​mu2.

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Given. m=1.5 kgm = 1.5\ \text{kg}m=1.5 kg, u=2.0 m/su = 2.0\ \text{m/s}u=2.0 m/s at x=1 mx = 1\ \text{m}x=1 m; F=2x2+3F = 2x^2 + 3F=2x2+3.

Plan. Integrate the force for the work, then use the work-energy theorem.

Step 1. Work. W=[23x3+3x]14=(1283+12)−(23+3)=51.0 JW = \Big[\tfrac23 x^3 + 3x\Big]_1^4 = \left(\tfrac{128}{3} + 12\right) - \left(\tfrac23 + 3\right) = 51.0\ \text{J}W=[32​x3+3x]14​=(3128​+12)−(32​+3)=51.0 J.

Step 2. Energy at the start. 12×1.5 kg×(2.0 m/s)2=3.0 J\tfrac12 \times 1.5\ \text{kg} \times (2.0\ \text{m/s})^2 = 3.0\ \text{J}21​×1.5 kg×(2.0 m/s)2=3.0 J.

Step 3. Speed. 12×1.5 kg×v2=3.0 J+51.0 J=54 J\tfrac12 \times 1.5\ \text{kg} \times v^2 = 3.0\ \text{J} + 51.0\ \text{J} = 54\ \text{J}21​×1.5 kg×v2=3.0 J+51.0 J=54 J, so v2=72 m2/s2v^2 = 72\ \text{m}^2/\text{s}^2v2=72 m2/s2 and v=8.49 m/sv = 8.49\ \text{m/s}v=8.49 m/s.

Answer: (a) 51.0 J51.0\ \text{J}51.0 J. (b) 8.49 m/s8.49\ \text{m/s}8.49 m/s.

Check. The force runs from 5 N5\ \text{N}5 N to 35 N35\ \text{N}35 N over 3 m3\ \text{m}3 m, so the work must lie between 15 J15\ \text{J}15 J and 105 J105\ \text{J}105 J. It does.

Watch out. Using F=35 NF = 35\ \text{N}F=35 N (its value at the end) for the whole way overstates the work: the force was smaller earlier.

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