M3 June 2013 (R) Q2
2. A particle of mass 4 kg is moving along the horizontal \(x\)-axis under the action of a single force which acts in the positive \(x\)-direction. At time \(t\) seconds the force has magnitude \(\left(1 + 3t^{\frac{1}{2}}\right)\) N.
When \(t = 0\) the particle has speed 2 m s\(^{-1}\) in the positive \(x\)-direction. Find the work done by the force in the interval \(0 \leqslant t \leqslant 4\) (7)
| Scheme | Marks |
|---|---|
| \(F = 1 + 3t^{\frac{1}{2}} = m\dfrac{\mathrm{d}v}{\mathrm{d}t} = 4\dfrac{\mathrm{d}v}{\mathrm{d}t}\) | B1 |
| \(4v = \displaystyle\int 1 + 3t^{\frac{1}{2}}\,\mathrm{d}t = t + 2t^{\frac{3}{2}}\ \ (+C)\) | M1A1 |
| \(v = \dfrac{1}{4}\left(t + 2t^{1.5}\right) + 2\) | A1 |
| \(t = 4,\ v = \dfrac{1}{4}(4 + 16) + 2 = 7\) (m s\(^{-1}\)) | A1ft |
| Work done = gain in KE \(= \dfrac{1}{2}mv^2 - \dfrac{1}{2}mu^2\) their \(v\) | M1 |
| \(= \dfrac{1}{2} \times 4 \times 7^2 - \dfrac{1}{2} \times 4 \times 2^2 = 90\) (J) | A1 |
| (7) | |
| (7 marks) |