A 15 kg box rests on a horizontal surface. What is the minimum horizontal force that is requiredto cause the box to begin to slide if the coefficient of static friction is 0.35? What is theacceleration of the system if a person pushed the box with a force of 90 N if the coefficient ofkinetic friction is 0.20? (3 sig. figs)

Sure, let's break this down into two parts.

Part 1: Minimum horizontal force to cause the box to slide

The force of static friction (Ff) can be calculated using the formula:

Ff = μN

where μ is the coefficient of static friction and N is the normal force. In this case, the normal force is equal to the weight of the box, which is its mass times the acceleration due to gravity (9.8 m/s^2). So,

N = mass * gravity N = 15 kg * 9.8 m/s^2 = 147 N

Now we can calculate the force of static friction:

Ff = μN Ff = 0.35 * 147 N = 51.45 N

So, the minimum horizontal force required to cause the box to begin to slide is 51.45 N.

Part 2: Acceleration of the system when pushed with a force of 90 N

When the box is pushed with a force of 90 N, the force of kinetic friction (Fk) comes into play. This can be calculated in a similar way to the force of static friction, but using the coefficient of kinetic friction (0.20) instead:

Fk = μkN Fk = 0.20 * 147 N = 29.4 N

The net force (Fnet) acting on the box is then the applied force (90 N) minus the force of kinetic friction:

Fnet = Applied force - Fk Fnet = 90 N - 29.4 N = 60.6 N

Finally, we can calculate the acceleration (a) of the box using Newton's second law (F = ma):

a = Fnet / mass a = 60.6 N / 15 kg = 4.04 m/s^2

So, the acceleration of the box when pushed with a force of 90 N is 4.04 m/s^2.