A) \[6.7\text{ }m/{{s}^{2}}\]
B) \[10\text{ }m/{{s}^{2}}\]
C) \[\text{3}\text{.3 }m/{{s}^{2}}\]
D) None of these
Correct Answer: A
Solution :
Let friction force = f |
\[F+f=ma\] ...(i) |
\[(F-f)R=I\alpha \] ...(li) |
From \[e{{q}^{n}}s,\] (i) and (ii), |
\[2F=ma+\frac{I\alpha }{R}\] |
Use \[\alpha =\frac{a}{R}\] (for pure rolling) |
\[2F=ma+\frac{Ia}{{{R}^{2}}}\] |
\[40=4a+\frac{0.02a}{{{(0.1)}^{2}}};\] \[a=\frac{40}{6}=6.7\,m/{{s}^{2}}\] |
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