A man pushes a block of ice across a frozen pond at a constant velocity. While the coefficients of static and kinetic friction f
1 answer:
Answer:
F₁> F₂
Explanation:
For this exercise Newton's second law is used, in the adjoint we can see the unapplied forces in this exercise.
Y axis y
N- W = 0
in this axis there is no movement
X axis
F -fr = m a
as they indicate that the velocity is consonant the acceleration is worth zero
F - fr = 0
friction force has the expression
fr = μ N
fr = μ mg
we substitute
F = μ m g
by the time the block is stopped the deferred force is
F₁ = μ_s m g
when it begins to move the force should decrease to
F₂ = μ_k k m g
as the static coefficient is greater than the dynamic coefficient
F₁> F₂
You might be interested in
To solve this problem it is necessary to apply the equations related to the Force of Friction and Energy.
By definition the friction force is defined as
Where,
Frictional Constant
N = Normal Force -> mg
At the same time we have the definition of the Energy, which can be
E = F*v
Where,
Force
v = Velocity.
Then replacing with our values we have that,
F = 0.5 *34
F = 17
The energy then would be,
E = f v
E = 17 * 0.30 = 5.1 W
Therefore the rate at which heat is generated is 5.1W