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:
Option B and C are True
Note: The attachment below shows the force diagram
Explanation:
The weight of the two blocks acts downwards.
Let the weight of the two blocks be W. Solving for T₁ and T₂;
w = T₁/cos 60° -----(1)
w = T₂/cos 30° ----(2)
equating (1) and (2)
T₁/cos 60° = T₂/cos 30°
T₁ cos 30° = T₂ cos 60°
T₂/T₁ = cos 30°/cos 60°
T₂/T₁ =1.73
Therefore, option a is false since T₂ > T₁
Option B is true since T₁ cos 30° = T₂ cos 60°
Option C is true because the T₃ is due to the weight of the two blocks while T₄ is only due to one block.
Option D is wrong because T₁ + T₂ > T₃ by simple summation of the two forces, except by vector addition.
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
The force tending to lift the load (vertical force) is equal to <u>22.5N.</u>
Why?
Since the boy is pulling a load (150N) with a string inclined at an angle of 30° to the horizontal, the total force will have two components (horizontal and vertical component), but we need to consider the given information about the tension of the string which is equal to 105N.
We can calculate the vertical force using the following formula:
Hence, we can see that <u>the force tending to lift the load</u> off the ground (vertical force) is equal to <u>22.5N.</u>
Have a nice day!