A block is released from rest on an inclined
plane and moves 3.1 m during the next 4.2 s.
The acceleration of gravity is 9.8 m/s^2
The block is 7 kg
The angle is26◦
What is the magnitude of the acceleration
of the block?
Answer in units of m/s^2
What is the coefficient of kinetic friction µk
for the incline?
To find the magnitude of the acceleration of the block, we can use the kinematic equation:
d = v₀t + (1/2)at²
Where:
- d is the displacement of the block (given as 3.1 m)
- v₀ is the initial velocity of the block (since it's released from rest, v₀ = 0)
- t is the time taken by the block (given as 4.2 s)
- a is the acceleration of the block (what we need to find)
Rearranging the equation, we can solve for a:
a = (2d) / t²
Substituting the given values, we get:
a = (2 * 3.1 m) / (4.2 s)²
a = 18.8 m/s² (rounded to one decimal place)
Therefore, the magnitude of the acceleration of the block is 18.8 m/s².
To find the coefficient of kinetic friction µk for the incline, we need to use the following equation relating the force of friction with the normal force:
F_friction = µk * F_normal
On an inclined plane, the normal force is given by:
F_normal = m * g * cos(θ)
Where:
- m is the mass of the block (given as 7 kg)
- g is the acceleration due to gravity (given as 9.8 m/s²)
- θ is the angle of the incline (given as 26°)
The force of gravity acting on the block can be given as:
F_gravity = m * g * sin(θ)
Since the block is moving and no longer at rest, the force of friction can be expressed as:
F_friction = m * a
Setting up the equations and substituting the given values, we can solve for µk:
m * a = µk * m * g * cos(θ)
µk = a / (g * cos(θ))
Substituting the given values, we get:
µk = 18.8 m/s² / (9.8 m/s² * cos(26°))
µk = 0.72 (rounded to two decimal places)
Therefore, the coefficient of kinetic friction µk for the incline is 0.72.