Un bloque de 15 kg, inicialmente en reposo, es empujado por una fuerza horizontal constante de 60N a lo largo de un pasillo. Si el coeficiente de fricción cinético entre el bloque y el piso es 0,2; encontrar la rapidez de la caja después de recorrer 6,5m.

5.15 m/s

Explanation:

First of all, we need to calculate the acceleration of the block.

The net force on the block is:

where:

is the applied force forward

is the force of friction backward, where

is the coefficient of kinetic friction

m = 15 kg is the mass of the block

is the acceleration due to gravity

Substituting,

According to Newton's second law of motion, the net force is equal to the product between the mass of the block (m) and its acceleration (a):

Solving for a, we find the block's acceleration:

The motion of the block is a uniformly accelerated motion, so we can use the following suvat equation to find the final speed of the block:

where

u = 0 is the initial speed

is the acceleration

s = 6.5 m is the distance covered

Solving for v, we find: