"an ice skater applies a horizontal force to a
20.-kilogram block on frictionless, level ice,
causing the block to accelerate uniformly at
1.4 meters per second2 to the right. after the
skater stops pushing the block, it slides onto a
region of ice that is covered with a thin layer
of sand. the coefficient of kinetic friction
between the block and the sand-covered ice is
0.28.
calculate the magnitude of the force of friction
acting on the block as it slides over the
sand-covered ice. [show all work, including the
equation and substitution with units.]
Using the Newton's Secound Law, which comes:
[tex]F_{at}=F_{r} \\ Nu=ma \\ N= \frac{20*1.4}{0.28} \\ \boxed {N=100N}[/tex]
If you notice any mistake in my english, please let me know, because i am not native.
The magnitude of the force of friction is about 55 N
[tex]\texttt{ }[/tex]
Further explanation
Newton's second law of motion states that the resultant force applied to an object is directly proportional to the mass and acceleration of the object.
[tex]\large {\boxed {F = ma }[/tex]
F = Force ( Newton )
m = Object's Mass ( kg )
a = Acceleration ( m )
Let us now tackle the problem !
[tex]\texttt{ }[/tex]
Given:
mass of block = m = 20 kg
acceleration of block = a = 1.4 m/s²
coeficient of kinetic friction = μ = 0.28
Asked:
magnitude of the force of friction = f = ?
Solution:
We will use following formula to solve this problem as follows:
[tex]f = \mu N[/tex]
[tex]f = \mu mg[/tex]
[tex]f = 0.28 \times 20 \times 9.8[/tex]
[tex]f = 54.88 \texttt{ Newton}[/tex]
[tex]f \approx 55 \texttt{ Newton}[/tex]
[tex]\texttt{ }[/tex]
Conclusion :
The magnitude of the force of friction is about 55 N
[tex]\texttt{ }[/tex]
Learn moreImpacts of Gravity : Effect of Earth’s Gravity on Objects : The Acceleration Due To Gravity : Newton's Law of Motion: Example of Newton's Law:
[tex]\texttt{ }[/tex]
Answer details
Grade: High School
Subject: Physics
Chapter: Dynamics
[tex]an ice skater applies a horizontal force to a 20.-kilogram block on frictionless, level ice, causin[/tex]
The only vertical forces are weight and normal force, and they balance since the surface is horizontal. The horizontal forces are the applied force (uppercase F) in the direction the block slides and the frictional force (lowercase f) in the opposite direction.
Apply Newton's 2nd Law in the horizontal direction:
ΣF = ma
F - f = ma
where f = µmg
F - µmg = ma
F = m(a +µg)
F = (20 kg)(1.4 m/s² + 0.28(9.8 m/s²)
F = 83 N
The vector diagram has been attached
[tex]An ice skater applies a horizontal force to a 20.-kilogram block on frictionless, level ice, causing[/tex]
Weight = (mass) x (gravity) .
The weight of the 20kg block is (20kg) x (9.8 m/s²) = 196 newtons .
Since the coefficient of kinetic friction between the block and
the sandy ice is 0.28, the frictional force opposite to the direction
in which the block is moving is
(0.28) x (weight) = (0.28 x 196 N) = 54.88 Newtons .
All that business about the skater pushing the block,
and the acceleration it has while she's pushing it, is
intriguing, but completely irrelevant to the actual question
and its solution.