Question

A motor of power 0.98kw drag a mass 100kg through 4m on a surface having cofficent of friction 0.3 . Time of that mass

Answer 1

power =force * displacement / time

given

P=0.98 kW

m=100 kg

s=4

mu= 0.3

( from s=at sqr)

F=mu ma (a = s/t sqr)

0.3*100*4/t sqr

120/t sqr

power =force * displacement / time

0.98 kW = 120/t sqr * 4/t

just calculate and u will get it

Answer 2

Given :

Power of motor , $P=0.98\ kW$ .

Mass of object , M = 100 kg .

Coefficient of friction , $\mu=0.3$ .

Initial velocity , u = 0 m/s .

To Find :

Time of that mass .

Solution :

Power is given by :

$P=\dfrac{Force \times Displacement }{time\ taken}\\\\P=\dfrac{Fd}{t}$ ...... 1

Now , frictional force F is given by :

$F=\mu Ma$      ........2

Here , a is acceleration .

Now , by equation of motion :

$d=ut+\dfrac{at^2}{2}$

Putting all given values in above equation .

We get :

$4=0\timest+\dfrac{at^2}{2}\\\\a=\dfrac{8}{t^2}$

Putting all given values in equation 2 , we get :

$P=\dfrac{Fd}{t}\\\\0.98\times 10^3=\dfrac{8\mu M}{t^3}\\\\0.98\times 10^3=\dfrac{8\times 0.3\times 100}{t^3}\\\\t=3\sqrt{\dfrac{8\times 0.3\times 100}{0.98\times 10^3}}\ s\\\\t=0.626\ s$

Therefore , time of that mass is 0.626 s .

Hence , this is the required solution .