Question

A block slides down a frictionless plane having an inclination of θ = 18.1°. The block starts from rest at the top, and the length of the incline is 2.85 m. (a) Draw a free-body diagram of the block.
(b) Find the acceleration of the block down the incline. (m/s^2 )
(c) Find its speed when it reaches the bottom of the incline. (m/s)

Answer 1

Answer:

(a)Look at the block free body diagram in the attached graphic

(b) a = 3.05 m/s²

(c) $v_{f} = 4.17 \frac{m}{s}$

Explanation:

Conceptual analysis

We apply Newton's second law:

∑F = m*a (Formula 1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Data:

θ = 18.1°

g = 9.81 m/s² : acceleration due to gravity  

μk = 0

X component of weight

We calculate the weight component parallel to the displacement of the box:

We define the x-axis in the direction of the inclined plane , 18.1° to the horizontal.

W= m*g :  Block weight

Wx= Wsin θ= m*g*sin  18.1°= (9.8 * sen18.1° )*m

Problem development

Look at the block free body diagram in the attached graphic

We apply the formula (1)

∑Fx = m*ax     , ax=a

Wx= m*a

(9.8*sin 18.1°) *m = m*a     : we divide by m on both sides of the equation

9.8*sin 18.1° = a

a = 3.05 m/s²

Kinematics of the block

Because the block moves with uniformly accelerated movement we apply the following formula:

vf²=v₀²+2*a*d Formula (2)

Where:  

d:displacement in meters (m)  

t : time in seconds (s)

v₀: initial speed in m/s  

vf: final speed in m/s  

a: acceleration in m/s²

Data

d= 2.85 m

v₀= 0

a = 3.05 m/s²

We replace the data in the formula (2)

vf²=v₀²+2*a*d

vf²=0+(2)(3.05)(2.85)

$v_{f} = \sqrt{(2)(3.05)(2.85)}$

$v_{f} = 4.17 \frac{m}{s}$