Question

A 5 kg block is sliding on a horizontal surface while being pulled by a child using a rope attached to the center of the block. The rope exerts a constant force of 28.2 N at an angle of \theta=θ = 30 degrees above the horizontal on the block. Friction exists between the block and supporting surface (with \mu_s=\:μ s = 0.25 and \mu_k=\:μ k = 0.12 ). What is the horizontal acceleration of the block?

Answer 1

Answer:

The horizontal acceleration of the block is 4.05 m/s².

Explanation:

The horizontal acceleration can be found as follows:

$F = m \cdot a$

$Fcos(\theta) - \mu_{k}N = m\cdot a$

$Fcos(\theta) - \mu_{k}[mg - Fsen(\theta)] = m\cdot a$  

$a = \frac{Fcos(\theta) - \mu_{k}[mg - Fsen(\theta)]}{m}$

Where:

a: is the acceleration

F: is the force exerted by the rope = 28.2 N

θ: is the angle = 30°

$\mu_{k}$: is the kinetic coefficient = 0.12

m: is the mass = 5 kg

g: is the gravity = 9.81 m/s²

$a = \frac{28.2 N*cos(30) - 0.12[5 kg*9.81 m/s^{2} - 28.2 N*sen(30)]}{5 kg} = 4.05 m/s^{2}$

Therefore, the horizontal acceleration of the block is 4.05 m/s².

I hope it helps you!