A 44 kg block lies on a horizontal frictionless surface. A horizontal force of 110 N is applied to the block. (a) Determine the
1 answer:
You might be interested in
Answer:
the ship's energy is greater than this and the crew member does not meet the requirement
Explanation:
In this exercise to calculate kinetic energy or final ship speed in the supply hangar let's use the relationship
W =∫ F dx = ΔK
Let's replace
∫ (α x³ + β) dx = ΔK
α x⁴ / 4 + β x = ΔK
Let's look for the maximum distance for which the variation of the energy percent is 10¹⁰ J
x (α x³ + β) = - K₀
= K₀ + x (α x³ + β)
Assuming that the low limit is x = 0, measured from the cargo hangar
Let's calculate
= 2.7 10¹¹ + 7.5 10⁴ (6.1 10⁻⁹ (7.5 10⁴) 3 -4.1 10⁶)
Kf = 2.7 10¹¹ + 7.5 10⁴ (2.57 10⁶ - 4.1 10⁶)
Kf = 2.7 10¹¹ - 1.1475 10¹¹
Kf = 1.55 10¹¹ J
In the problem it indicates that the maximum energy must be 10¹⁰ J, so the ship's energy is greater than this and the crew member does not meet the requirement
We evaluate the kinetic energy if the System is well calibrated
W = x F₀ = –K₀
= K₀ + x F₀
We calculate
= 2.7 10¹¹ -7.5 10⁴ 3.5 10⁶
= (2.7 -2.625) 10¹¹
= 7.5 10⁹ J
To solve this problem we will apply the concepts related to the balance of forces. We will decompose the forces in the vertical and horizontal sense, and at the same time, we will perform summation of torques to eliminate some variables and obtain a system of equations that allow us to obtain the angle.
The forces in the vertical direction would be,
The forces in the horizontal direction would be,
The sum of Torques at equilibrium,
The maximum friction force would be equivalent to the coefficient of friction by the person, but at the same time to the expression previously found, therefore
Replacing,
Therefore the minimum angle that the person can reach is 46.9°
Answer:
"1155 N" is the appropriate solution.
Explanation:
Given:
Acceleration,
Forces resisting motion,
Mass,
By using Newton's second law, we get
⇒
Or,
⇒
By putting the values, we get
⇒
⇒
⇒