Problem Statement
A constant force $F = 10\,\text{N}$ acts on a block of mass $m = 2\,\text{kg}$ at an angle of $\theta = 0^{\circ}$ to the horizontal, moving it a distance $d = 4\,\text{m}$ along a smooth horizontal floor. The block starts from rest. Find (a) the work done by the force, (b) the speed of the block after it has moved this distance using the work–energy theorem, and (c) the average power delivered if the motion takes $t = 2\,\text{s}$.
Given Information
- $F = 10\,\text{N}$ — applied force
- $m = 2\,\text{kg}$ — mass of the block
- $d = 4\,\text{m}$, $\theta = 0^{\circ}$, $t = 2\,\text{s}$
- Floor is smooth (frictionless); starts from rest
Physical Concepts & Formulas
Work done by a constant force is $W=Fd\cos\theta$, only the component of force along the displacement contributes. The work–energy theorem states the net work equals the change in kinetic energy. Average power is the work done divided by the time taken.
- $W = Fd\cos\theta$ — work by a constant force
- $W_{net} = \Delta KE = \tfrac12 mv^{2} – \tfrac12 mv_0^{2}$
- $v = \sqrt{\dfrac{2W}{m}}$ — from rest
- $P_{avg} = \dfrac{W}{t}$ — average power
Step-by-Step Solution
Step 1 — Compute the work done by the force: Only the component $F\cos\theta$ along the displacement does work.
$$ W = 10\times4\times\cos0^{\circ} = 40\,\text{J} $$
Step 2 — Apply the work–energy theorem: On a smooth floor this is the only horizontal work, so it equals the gain in kinetic energy.
$$ W = \tfrac12 mv^{2} – 0 $$
Step 3 — Solve for the final speed: Rearrange the work–energy relation for $v$.
$$ v = \sqrt{\frac{2W}{m}} = \sqrt{\frac{2\times40}{2}} = 6.32\,\text{m/s} $$
Step 4 — Compute the average power: Divide the total work by the time interval.
$$ P_{avg} = \frac{40}{2} = 20\,\text{W} $$
Worked Calculation
$$ W = 10\times4\cos0^{\circ} = 40\,\text{J},\quad v = \sqrt{2W/m} = 6.32\,\text{m/s} $$
Answer
$$ \boxed{W = 40\,\text{J},\ v = 6.32\,\text{m/s},\ P_{avg} = 20\,\text{W}} $$
All the work done by the force becomes kinetic energy because the floor is frictionless and the motion is horizontal.
Physical Interpretation
The work–energy theorem bypasses the need to track acceleration with time — it links force and distance directly to speed. If friction were present, part of this work would be dissipated as heat and the final speed would be lower.
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