Question

A bodypulls 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 does the boy do on the block?$Takeg=9.8m/s^2$.

• 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 $F$ $\sin {45}^0.$

Thus the net force by the block on the ground $=\left(mg-F\sin {45}^0\right)$

So$,$ the frictional force acting on the block is $=\mu \left(mg-F\sin {45}^0\right)$

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

So$,$ $F\cos {45}^0=\mu \left(mg-F\sin {45}^0\right)$

$\Rightarrow F\cos {45}^0+\mu F\sin {45}^0=\mu mg$

$\Rightarrow \left(F\cos {45}^0\right)\left(1+\mu \tan {45}^0\right)=\mu mg$

$\Rightarrow \left(F\cos {45}^0\right)\left(1+\mu \right)=\mu mg$

$\Rightarrow F\cos {45}^0=\mu mg/\left(1+\mu \right)$

Now the work done on the block is $=\left(F\right)\left(d\right)\left(\cos {45}^0\right)$

$=\left(F\cos {45}^0\right)\left(d\right)$

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

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

$=196/1.2$

$=163.33J$

Hence,

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