Question

9) A string exerts a force of 58 N on a box at an angle of 34 degrees from the horizontal. The box has a mass of 4 kg

Answer 1

The acceleration of the box that has a mass of 4 kg and a coefficient of kinetic friction of 0.2, when moved by an exerted force of 58 N, is 10.06 m/s².    

The acceleration of the box can be calculated with the sum of the forces acting on the x-direction:

$\Sigma F_{x} = ma$

We will not take into account the forces acting on the y-direction since we need to find the acceleration of the box when it is moving (x-direction).

$T_{x} - F_{\mu} = ma$    (1)

Where:

• $T_{x}$: is the horizontal component of the tension force (T) exerted by the string on the box = T*cos(θ) 

• θ: is the angle (from the horizontal) = 34°

• $F_{\mu}$: is the force due to the kinetic friction = -μN (the minus sign is because it is in the opposite direction of motion)  

• m: is the mass of the box = 4 kg

• a: is the acceleration =?

• μ: is the coefficient of kinetic friction = 0.2

• N: is the normal force = mg

• g: is the acceleation due to gravity = 9.81 m/s²

From equation (1) we have:

$Tcos(34) - \mu N = ma$  

$Tcos(34) - \mu mg = ma$  

Hence, the acceleration is:

$58N*cos(34) - 0.2*4 kg*9.81 m/s^{2} = 4 kg*a$  

$a = \frac{58N*cos(34) - 0.2*4 kg*9.81 m/s^{2}}{4 kg} = 10.06 m/s^{2}$

Therefore, the acceleration of the box when the box is moving is 10.06 m/s².

You can learn more about forces here:

• brainly.com/question/2008782?referrer=searchResults

• brainly.com/question/24611730?referrer=searchResults

 

I hope it helps you!