A body of mass M rests on a horizontal plane having coefficient of friction μ. At t = 0 a horizontal force F→ it applied that varies with time F→(t)=F→0t, where F→0 is a constant vector. The time instant t0 at which motion starts and distance moved in t seconds will be:

To start motion, F0t0≥μMg⇒t0=μMgF0Now Mdvdt=F0t−μMg=F0t−F0t0⇒M∫0v dv=∫t0t F0t−t0dt⇒v=F0t−t022MTherefore, dxdt=F0t−t022M⇒x=∫dx=∫t0t F02t−t02dt⇒x=F06Mt−t03

A body of mass M rests on a horizontal plane having coefficient of friction μ. At t = 0 a horizontal force F→ it applied that varies with time F→(t)=F→0t, where F→0 is a constant vector. The time instant t0 at which motion starts and distance moved in t seconds will be:

To start motion, F0t0≥μMg⇒t0=μMgF0Now Mdvdt=F0t−μMg=F0t−F0t0⇒M∫0v dv=∫t0t F0t−t0dt⇒v=F0t−t022MTherefore, dxdt=F0t−t022M⇒x=∫dx=∫t0t F02t−t02dt⇒x=F06Mt−t03

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A body of mass M rests on a horizontal plane having coefficient of friction $μ$ . At t = 0 a horizontal force $\vec{F}$ it applied that varies with time $\vec{F} (t) = {\vec{F}}_{0} t$ , where ${\vec{F}}_{0}$ is a constant vector. The time instant t₀ at which motion starts and distance moved in t seconds will be:

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$\frac{Mg}{{μF}_{0}} and \frac{F_{0}}{M} {(t - t_{0})}^{2}$

$\frac{μMg}{F_{0}} and \frac{1}{6} \frac{F_{0}}{M} {(t - t_{0})}^{3}$

$\frac{{μF}_{0}}{Mg} and \frac{F_{0}}{M} {(t - t_{0})}^{2}$

$\frac{F_{0}}{Mg} and \frac{F_{0}}{2 Mg} {(t - t_{0})}^{2}$

answer is B.

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Detailed Solution

To start motion, $F_{0} t_{0} \geq μMg \Rightarrow t_{0} = \frac{μMg}{F_{0}}$
Now $M \frac{dv}{dt} = F_{0} t - μMg = F_{0} t - F_{0} t_{0}$
$\Rightarrow M \int_{0}^{v} dv = \int_{t_{0}}^{t} F_{0} (t - t_{0}) dt \Rightarrow v = \frac{F_{0} {(t - t_{0})}^{2}}{2 M}$
Therefore, $\frac{dx}{dt} = \frac{F_{0} {(t - t_{0})}^{2}}{2 M}$
$\Rightarrow x = \int dx = \int_{t_{0}}^{t} \frac{F_{0}}{2} {(t - t_{0})}^{2} dt \Rightarrow x = \frac{F_{0}}{6 M} {(t - t_{0})}^{3}$