Question

58. Find the magnitude of the force applied to a block
of mass 5 kg at rest if it moves 36 m in the first 6
seconds. Neglect the force of friction. ​

Answer 1

Answer: The magnitude of force applied = 10N

Using the formula

s = ut + 1/2at^2

a = 2m/s^2

Now force applied

F = ma = 5 x 2 = 10N

Answer 2

$\bf{\underline{\underline{Given :}}}$

$\mathtt{\bigstar\: Initial\:velocity\:(u) = 0\:ms^{-1}\:(rest)}$

$\mathtt{\bigstar\: displacement\:(s) = 36\: m}$

$\mathtt{\bigstar\: time\:(t) = 6\: s}$

$\mathtt{\bigstar\: mass (m) = 5\:kg}$

$\bf{\underline{\underline{Let\: us\: find\: out \: acceleration}}}$

$\mathtt{Using\:\: 2nd\:\: eqn\:\: of\:\: motion}$

$\fbox{\mathtt{\bigstar\:\:s = ut + \Large{\frac{1}{2}}} \mathtt{at^{2} }}$

$\mathtt{\rightarrow\: 36 = 0 × 6 + \Large{\frac{1}{2}}} \mathtt{ × a × (6)^{2}}$

$\mathtt{\rightarrow\: 36 = 0 + \Large{\frac{1}{\cancel{2}}}} \mathtt{× a × \cancel{36}}$

$\mathtt{\rightarrow\: 36 = a × 18}$

$\mathtt{\rightarrow\: a = \Large{\frac{36}{18}}}$

$\mathtt{\rightarrow\: a = 2\:ms^{-2}}$

$\fbox{\mathtt{\red{Force = mass × acceleration}}}$

$\mathtt{\rightarrow\: F = 5\:kg × 2\:ms^{-2}}$

$\fbox{\Large{\mathtt{\green{\rightarrow\: F = 10\:N}}}}$

Therefore, 10N force is applied on the block