Question

A body pulls a $5kg$ block $20m$ along a horizontal surface at a constant speed with a force directed ${45}^o$ above the horizontal. If the coefficient of kinetic friction is $0.2$, how much work does the boy do on the block? Take $g=9.8m/s^2$

• A. $32.5J$
• B. $113.5J$
• C. $163.3J$
• D. $25.00J$

Answer 1

The correct option is C $163.3J$

Here$,$

Mass of the block$,m=5kg$

Coefficient of friction$,\mu =0.2$

Distance traveled$,d=20m$

Let the force be $F.F$ makes an angle ${45}^0$ with the horizontal.

The vertical component of the applied force is $Fsin45.$

Thus the net force by the block on the ground $=(mg–Fsin45)$

So$,$ the frictional force acting on the block is $=\mu (mg–Fsin45)$

The block moves at a constant speed. That means the horizontal component of the force just nullifies the frictional force$.$

So$,Fcos45=\mu (mg–Fsin45)$

$=>Fcos45+\mu Fsin45=\mu mg$

$=>\left(Fcos45\right)(1+\mu tan45)=\mu mg$

$=>\left(Fcos45\right)(1+\mu )=\mu mg$

$=>Fcos45=\mu mg/(1+\mu )$

Now the work done on the block is $4=\left(F\right)\left(d\right)\left(cos45\right)$

$=\left(Fcos45\right)\left(d\right)$

$=\left[\mu mg/\right(1+\mu \left)\right]\left(d\right)$

$=\left(0.2\right)\left(5\right)\left(9.8\right)\left(20\right)/(1+0.2)$

$=196/1.2$

$=163.33J$

Hence,

option $\left(C\right)$ is correct answer.