Question

A 55 kg box rests on a horizontal surface. the coefficient of static friction between the box and the surface is 0.3 and coefficient of kinetic friction is 0.2 what horizontal force must be applied to the

Answer 1

Horizontal force must be applied to load

Answer 2

Correct Question:

A 55-kg box rests on a horizontal surface. The coefficient of static friction between the box the surface is 0.30, and the coefficient of kinetic friction is 0.20. What horizontal force must be applied to the box to cause it to start sliding along the surface?

Calculation:

In order to move the box , we have to apply a force equal to the critical static friction such that the box starts moving.

So, let critical static friction be f ;

$\rm{ \therefore \: f = \mu \times N}$

Now , in this case , the normal reaction will be equal to the weight of the box.

$\rm{ = > \: f = \mu \times mg}$

$\rm{ = > \: f = 0.3 \times (55 \times 10)}$

$\rm{ = > \: f = 3 \times 55 }$

$\rm{ = > \: f = 165 \: newton }$

So, we have to apply a force of 165 N in order to move the object.

So, final answer is:

$\boxed{\bold{\: F_{applied}= 165 \: newton }}$