Answer:
v = 126 m / s
Explanation:
Let's analyze this exercise a little, they give us the thrust that is the applied force and the time that it lasts, and they ask us for the final speed, so we can use the Impulse ratio and the variation of the amount of movement
I = F t = Dp
F t = pf -p₀
Now let's use Newton's second law to find the net thrust
F = E - fr
the friction force has the formula
fr = μ N
let's write Newton's second law on the y-axis
N-W = 0
N = W
we substitute
fr = μ mg
we look for the net out
F = 200 - μ mg
With the skater starting from rest, the initial speed is zero (vo = 0)
we substitute
(200 - very m g) t = m v
v = (200 µm - very g) t
let's calculate
v = (200/75 - 0.10 9.8) 75
v = 126 m / s
Two workers are sliding 330 kg crate across the floor. One worker pushes forward on the crate with a force of 430 N while the other pulls in the same direction with a force of 330 N using a rope connected to the crate. Both forces are horizontal, and the crate slides with a constant speed. What is the crate's coefficient of kinetic friction on the floor?
Answer:
Coefficient of kinetic friction = 0.235
Explanation:
Given:
Mass of crate = 330 kg
1st force = 430 N
2nd force = 330 N
Find:
Coefficient of kinetic friction.
Computation:
We know that, velocity is constant.
So, acceleration (a) = 0
So, net force (f) = 430 N + 330 N
Net force (f) = 760 N
F = μmg
μ = f / mg [∵ g = 9.8]
μ = 760 / [330 × 9.8]
μ = 760 / [3,234]
μ = 0.235
Coefficient of kinetic friction = 0.235