Question

A 50.0 N box is at rest on a horizontal surface. The coefficient of static friction between the box and the
surface is 0.50, and the coefficient of kinetic friction is 0.30. A horizontal 20.0 N force is then exerted
on the box. The magnitude of the acceleration of the box is most nearly​

Answer 1

Answer:

Acceleration will be $a=0m/sec^2$

Explanation:

We have given that a 50 N box is at rest

Coefficient of static friction = 0.5

And coefficient of kinetic friction = 0.3

Force exerted on the box = 20 N

Static friction force = $coefficint\ of\ static\ friction\times 50=0.5\times 50=25N$

As the static friction force is greater than the force exerted force so box will not move and so acceleration will be $a=0m/sec^2$

Answer 2

Acceleration is the rate of change of the velocity of an object with respect to time. The acceleration of  50.0 N box is at rest on a horizontal surface is zero.

Acceleration:

It is the rate of change of the velocity of an object with respect to time.

Given Here ,

50 N box is at rest

Coefficient of static friction = 0.5

Coefficient of kinetic friction = 0.3

Force exerted on the box = 20 N

Static friction force formula

$\rm \bold { Fs = \mu _s N}$

$\rm \bold { Fs = 0.5 \times 50 }\\\\\rm \bold { Fs = 25}$

The Static friction force is grater than the force exerted on the box.

Hence we can conclude that the acceleration of  50.0 N box is at rest on a horizontal surface is zero.

To know more about acceleration, refer to the link:

https://brainly.com/question/24372530