Answer:
fr = ½ m v₀²/x
Explanation:
This exercise the body must be on a ramp so that a component of the weight is counteracted by the friction force.
The best way to solve this exercise is to use the energy work theorem
W = ΔK
Where work is defined as the product of force by distance
W = fr x cos 180
The angle is because the friction force opposes the movement
Δk = –K₀
ΔK = 0 - ½ m v₀²
We substitute
- fr x = - ½ m v₀²
fr = ½ m v₀²/x
Answer:
A) F = - 8.5 10² N, B) I = 21 N s
Explanation:
A) We can solve this problem using the relationship of momentum and momentum
I = Δp
in this case they indicate that the body rebounds, therefore the exit speed is the same in modulus, but with the opposite direction
v₀ = 8.50 m / s
v_f = -8.50 m / s
F t = m v_f -m v₀
F =
let's calculate
F =
F = - 8.5 10² N
B) let's start by calculating the speed with which the ball reaches the ground, let's use the kinematic relations
v² = v₀² - 2g (y- y₀)
as the ball falls its initial velocity is zero (vo = 0) and the height upon reaching the ground is y = 0
v =
calculate
v =
v = 14 m / s
to calculate the momentum we use
I = Δp
I = m v_f - mv₀
when it hits the ground its speed drops to zero
we substitute
I = 1.50 (0-14)
I = -21 N s
the negative sign is for the momentum that the ground on the ball, the momentum of the ball on the ground is
I = 21 N s