Question

A motor of power $1.98$ kW drags a mass $100$ kg through $4$m on a surface having coefficient of friction $0.3$. Time of that mass is?

• A. $0.4$ sec
• B. $0.8$ sec
• C. $1.2$ sec
• D. $0.6$ sec

Answer 1

The correct option is A $0.4$ sec

$power=\frac{force\times displacement}{time}$

given

$P=0.98kW$

$m=100kg$

$s=4$

$mu=0.3$

$(froms={at}^2)$

$F=mu\times ma\times (a=\frac{s}{t^{2}r})$

$=0.3\ast 100\ast \frac{4}{t^{2}r}=\frac{120}{t^{2}r}$

$power=\frac{force\times displacement}{time}$

$0.98kW=\frac{120}{t^{2}r}\times \frac{4}{t}$

$=0.4sec$