Question

A 4 kg box slidesdown a vertical wall with constant speed while a person pushes it up at an angle 45 degree with the vertical .If the coefficient of friction between the block and the wall is 0.41 then the magnitude of the force applied by the person is close to​

Answer 1

Given:

Mass of the box (m) = 4kg

Person pushes up the box at an angle $45^o$

Coefficient of friction ($\mu$) = 0.41

To Find:

Magnitude of applied force on the block by the person.

Solution:

Here, angle of force = $45^o$

Let, the force be, F

The component of the force are $F\cos 45^o$ and $F\sin 45^o$.

Since, we know that frictional force =  $Normal\hspace{1mm} force(N)\times \mu$

Here, normal force to the block is $F\sin 45^o$

∴Frictional force = $F\sin 45^o\times \mu$

And in the box gravitational force is also acting, hence, $F_g=mg$

In the question it says that the speed is constant, so the net force is also constant

∴$F_{net} = 0$

$=> mg-F\cos 45^o-F\sin 45^o\times \mu=0$ (∵$F\cos 45^o$ and frictional force is in the opposite direction to the gravitational force)

$=> 4\times 9.8-F\frac{1}{\sqrt{2} } -F\frac{1}{\sqrt{2} }\times 0.41=0$

$=> 4\times 9.8-F\frac{1}{\sqrt{2} } (1+0.41)=0$

$=>F\times 0.99 =39.2$

$=>F \approx 39.2 N$

∴The magnitude of applied force by the man is approx. 39.2 N.