Question

Example 7.1 Mr. Clean 50.0N A man cleaning a floor pulls a vacuum cleaner with a force of magnitude F= 50.0 N at an angle of 30.0° with the horizontal (Fig. 7.5). Calculate the work done by the force on the vacuum cleaner as the vac- uum cleaner is displaced 3.00 m to the right. 30.00

****** PLEASE ANSWER BOTH THE QUESTIONS *********

Answer 1

Work is the scalar product of force and displacement.

W=F. S

W=FScos$\Theta$

Here theta is the angle between force and displacement.

Given, F=50N

S=3m

$\Theta$=30°

Therefore W=50*3*cos30°=129.90J

Here weight and its normal reactions are perpendicular to displacement so work done by these forces will be zero.

In the case of truck,

Weight of truck, mg, is acting downward and crane is applying an upward force F.so the resultant force is F-mg.It displaces truck through a distance of 2m so work done by the crane

W= (F-mg)*2

F=31kN

mg=3000*9.8=29.4kN

Therefore W=(31*10^3-29.4*10^3)*2=3.2kJ.

Speed of truck after 2m.

Here a constant force F-mg=31-29.4=1.6kN is acting over 2m.The work done will convert in to kinetic energy at 2m.It srats with zero speed so W=(1/2)mv^2

3.2*10^3=(1/2)3000*v^2

V=1.46m/s.