What was the average force applied to the box in the direction of motion?

[kooaznboi1088]

A 7 kg box increases its speed form 2 to 4 m/s over a horizontal distance of 5 meters. coefficient fo kinetic friction is .1. what was the average force applied tothe box int he direction of the motion?

 

[Hootenanny]

Could you show some of your working or ideas?

 

[quicknote]

To solve this problem, we can use the equation F = ma, where F is the force applied, m is the mass of the box, and a is the acceleration. We can also use the formula for kinetic friction, Fk = μk * N, where μk is the coefficient of kinetic friction and N is the normal force.

First, we need to find the acceleration of the box. We can use the equation v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, and s is the distance traveled. Plugging in the given values, we get:

4^2 = 2^2 + 2a*5

16 = 4 + 10a

10a = 12

a = 1.2 m/s^2

Next, we can calculate the normal force using the formula N = mg, where m is the mass of the box and g is the acceleration due to gravity (9.8 m/s^2). Plugging in the given mass of 7 kg, we get:

N = 7 kg * 9.8 m/s^2 = 68.6 N

Now, we can use the formula for kinetic friction to find the force of friction:

Fk = μk * N = 0.1 * 68.6 N = 6.86 N

Finally, we can use the equation F = ma to find the average force applied to the box:

F = ma = 7 kg * 1.2 m/s^2 = 8.4 N

Therefore, the average force applied to the box in the direction of motion is 8.4 N.