A force \( (3x^2 + 2x - 5) \, \text{N} \) displaces a body from \( x = 2 \, \text{m} \) to \( x = 4 \, \text{m} \). The work done by this force is _________ J.
A force \( (3x^2 + 2x - 5) \, \text{N} \) displaces a body from \( x = 2 \, \text{m} \) to \( x = 4 \, \text{m} \). The work done by this force is _________ J.
Correct Answer: 58
Approach Solution - 1
To find the work done by the force \( F(x) = 3x^2 + 2x - 5 \) as it displaces a body from \( x = 2 \) m to \( x = 4 \) m, we use the work integral:
\( W = \int_{2}^{4} F(x) \, dx = \int_{2}^{4} (3x^2 + 2x - 5) \, dx \)
First, compute the indefinite integral:
\(\int (3x^2 + 2x - 5) \, dx = \int 3x^2 \, dx + \int 2x \, dx - \int 5 \, dx = x^3 + x^2 - 5x + C\)
Now, apply the limits of integration from 2 to 4:
\(W = [x^3 + x^2 - 5x]_{2}^{4} = [(4^3 + 4^2 - 5 \times 4) - (2^3 + 2^2 - 5 \times 2)]\)
Calculate each term:
\(4^3 = 64, \, 4^2 = 16\)
\(2^3 = 8, \, 2^2 = 4\)
Substitute these into the equation:
\(= (64 + 16 - 20) - (8 + 4 - 10)\)
\(= 60 - 2\)
\(= 58 \, \text{J}\)
The work done by the force is \(58\) J, which fits within the provided range (58,58).
Approach Solution -2
Step 1 — Write the integral for work
The work done by a force that varies with position is the definite integral of the force over the path:
\[ W=\int_{x=2}^{x=4} \big(3x^{2}+2x-5\big)\,dx. \]
Step 2 — Find an antiderivative (indefinite integral)
Integrate term-by-term:
- \(\displaystyle \int 3x^{2}\,dx = x^{3}\)
- \(\displaystyle \int 2x\,dx = x^{2}\)
- \(\displaystyle \int (-5)\,dx = -5x\)
So an antiderivative is
\(F(x)=x^{3} + x^{2} - 5x\)
Step 3 — Evaluate the definite integral
Compute \(F(4)\) and \(F(2)\):
\[ \begin{aligned} F(4) &= 4^{3} + 4^{2} - 5\cdot 4 = 64 + 16 - 20 = 60,\\[6pt] F(2) &= 2^{3} + 2^{2} - 5\cdot 2 = 8 + 4 - 10 = 2. \end{aligned} \]
Now subtract:
\[ W = F(4) - F(2) = 60 - 2 = 58 \]
So the work done is 58 J.
Units & physical meaning
Force is in newtons (N), displacement in metres (m). The result of integration is in joules (J).
Work done = 58 J.
This positive value means the force does net positive work on the object from 2 m to 4 m.
Optional check — average force
The average force is
\(\overline{F} = \dfrac{1}{4-2}\int_{2}^{4} F(x)\,dx = \dfrac{58}{2} = 29\ \text{N}\).
Multiplying by displacement gives \(29 \times 2 = 58\) J, consistent with the integral.
Common mistakes to avoid
- Forgetting to evaluate both limits (upper minus lower).
- Incorrect signs in the antiderivative.
- Omitting units — always state joules (J).
Quick recap
Work done \(=\displaystyle\int_{2}^{4} (3x^{2}+2x-5)\,dx = \boxed{58\ \text{J}}\).
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