Answer to Question #261647 in Physics for Jason

Question

A block of weight 7.0N rest on a level floor. The frictional force between the block and the floor is 1.0N. A horizontal force of 1.4N is used to pull the block for 4seconds. What is the velocity of the block after this time?

Answer 1

The total force on the block is:

$$F = 1.4N-1.0N = 0.4N$$

where "-" sign is because the frictional force and pulling force are opposite.

Let the mass of the block be:

$$m = \dfrac{7N}{9.8N/kg} = \dfrac{7}{9.8}kg$$

where $9.8N/kg$ is the gravitational acceleration.

Then, according to the second Newton's law, the acceleration of the block is:

$$a = \dfrac{F}{m}$$

The velocity after time $t=4s$ is given by the kinematic equation:

$$v = at = \dfrac{Ft}{m}\\ v = \dfrac{0.4N\cdot 4s}{7/9.8kg} \approx 2.2 m/s$$

Answer. 2.2 m/s.