A 1.00x102 kg crate is being pushed across a horizontal floor by a force pthat makes an angle of 30 degrees below the horizontal. the coefficient of kinetic friction is 0.200. what should be the magnitude of p so that the net work done by it and the kinetic frictional force is zero?

Respuesta :

Answer:

226.6 N

Explanation:

The work done by the push is given by:

[tex]W_p = p d cos \theta[/tex]

where

p is the magnitude of the push

d is the displacement of the crate

[tex]\theta=30^{\circ}[/tex] is the angle between the direction of the push and the displacement of the crate

On the contrary, the work done by the frictional force is:

[tex]W_f = - F_f d[/tex]

where

[tex]F_f = \mu mg[/tex] is the frictional force, with m=1.00x102 kg being the mass of the crate ang g=9.81 m/s^2 the gravitational acceleration

d is the displacement of the crate

the negative sign is due to the fact that the frictional force acts in the opposite direction to the displacement of the crate.

The net work must be zero, so:

[tex]W_p + W_f =0[/tex]

Substituting the previous equations into this one, we find:

[tex]pd cos \theta - \mu d mg=0[/tex]

which we can solve to find p:

[tex]p=\frac{\mu mg}{cos \theta}=\frac{(0.200)(1.00\cdot 10^2 kg)(9.81 m/s^2)}{cos 30^{\circ}}=226.6 N[/tex]