Exercise 7.11
A kg block is moving on horizontal floor with negligible friction. As shown in the Fig. , a variable force is applied on the block in its direction of motion from its position at m till m. If the block had a kinetic energy of J when it was at m, find the block’s speed (i) at m, and (ii) at m. Does the block have negative acceleration in any portion of its motion?
NCERT’s answer
$\displaystyle 6$ m \(\displaystyle s^{- 1}\); √$\displaystyle 66$ m \(\displaystyle s^{- 1}\); No
(i) Speed at $\displaystyle 0$ m = \(\displaystyle 6\ \text{m s}^{-1} \); (ii) speed at $\displaystyle 4$ m = \(\displaystyle \sqrt{66} \approx 8.1\ \text{m s}^{-1} \); and no, the block never has negative acceleration.
(i) Speed at $\displaystyle 0$ m — from \(\displaystyle K = \frac{1}{2}mv^{2} \):
\(\displaystyle 180\ \text{J} = \frac{1}{2} \times 10.0\ \text{kg} \times v^{2} \)
\(\displaystyle v^{2} = 36\ \text{m}^{2}\,\text{s}^{-2} \), so \(\displaystyle v = 6\ \text{m s}^{-1} \)
(ii) Speed at $\displaystyle 4$ m — the work done by a varying force is the area under the force-displacement graph (Fig. $\displaystyle 7.37$). The graph is a trapezium: the force climbs to $\displaystyle 50$ N by $\displaystyle 1$ m, stays at $\displaystyle 50$ N up to $\displaystyle 3$ m, then falls back to $\displaystyle 0$ at $\displaystyle 4$ m.
Area \(\displaystyle = \frac{1}{2} \times (4\ \text{m} + 2\ \text{m}) \times 50\ \text{N} = 150\ \text{J} \)
By the work-energy theorem, \(\displaystyle K_{\text{final}} = 180\ \text{J} + 150\ \text{J} = 330\ \text{J} \)
\(\displaystyle 330\ \text{J} = \frac{1}{2} \times 10.0\ \text{kg} \times v^{2} \Rightarrow v^{2} = 66\ \text{m}^{2}\,\text{s}^{-2} \)
\(\displaystyle v = \sqrt{66} \approx 8.1\ \text{m s}^{-1} \)
Negative acceleration? No. The applied force stays in the direction of motion everywhere between $\displaystyle 0$ m and $\displaystyle 4$ m — it never becomes negative — and friction is negligible.
Between $\displaystyle 3$ m and $\displaystyle 4$ m the force gets smaller, so the block gains speed more slowly, but it is still speeding up. A smaller forward force is not a backward force.