Questions & Answers

Question

Answers

$\left( A \right)0.25$

$\left( B \right)0.5$

$\left( C \right)2$

$\left( D \right)4$

Answer
Verified

${{F}_{static}}={{\mu }_{static}}\times mass\times g,{{F}_{friction}}={{\mu }_{friction}}\times mass\times g,Total\,force={{F}_{static}}-{{F}_{friction}}$ and $Total\,force=ma$ where m is the mass and a is the acceleration. Also, ${{F}_{friction}}$ is the frictional force, ${{F}_{static}}$ is static force. And g is the force of gravity.

As it is given that the mass of the object to which the static force is applied is given as 4 kg. The static force applied here has the value of 0.8 along with the gravitational force as $g=10m{{s}^{-2}}$. To find the force here, we will use formula ${{F}_{static}}={{\mu }_{static}}\times mass\times g$ and substitute the values in it. This result into,

$\begin{align}

& {{F}_{static}}={{\mu }_{static}}\times mass\times g \\

& \Rightarrow {{F}_{static}}=0.8\times 4\times 10N \\

& \Rightarrow {{F}_{static}}=8\times 4N \\

& \Rightarrow {{F}_{static}}=32N \\

\end{align}$

As it is given that the mass of the object to which the frictional force is applied is given as 4 kg. The frictional force applied here has the value of 0.6 along with the gravitational force as $g=10m{{s}^{-2}}$ . To find the force here, we will use formula ${{F}_{friction}}={{\mu }_{friction}}\times mass\times g$ and substitute the values in it. This result into,

$\begin{align}

& {{F}_{friction}}={{\mu }_{friction}}\times mass\times g \\

& \Rightarrow {{F}_{friction}}=0.6\times 4\times 10N \\

& \Rightarrow {{F}_{friction}}=6\times 4N \\

& \Rightarrow {{F}_{friction}}=24N \\

\end{align}$

Now, the total force here will be $Total\,force={{F}_{static}}-{{F}_{friction}}$ and $Total\,force=ma$ so, we have

$\begin{align}

& {{F}_{static}}-{{F}_{friction}}=ma \\

& \Rightarrow 32-24=\left( 4 \right)a \\

& \Rightarrow a=\dfrac{8}{4} \\

& \Rightarrow a=2 \\

\end{align}$

Therefore, the acceleration is $2m{{s}^{-2}}$.

To solve these types of questions we will learn the following points effectively,

(1) Static force is always greater than frictional force or kinetic force.

(2) The force should be calculated in Newton as Newton is the SI unit of force.

(3) The gravitational force is present on all the objects that are present around us. The general value of this force is $9.8m{{s}^{-2}}$ but since, we are given the value as $g=10m{{s}^{-2}}$ in this question, this is why we have used it. Otherwise, if the value of g is not specifically given in the question so, always use g as $9.8m{{s}^{-2}}$.