2. You decide to lift a 25 kg box to a height of 3 meters. How much work will you do while moving
1 answer:
Answer:
W = 735.75[J]
Explanation:
Work is defined as the product of force by distance. Therefore we can use the following equation.
where:
W = work [J] (units of Joules)
F = force [N] (units of Newtons)
d = distance = 3 [m]
But first, we must determine the force that is equal to the product of mass by gravity (weight of the body).
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Answer:
a = 30.832 ft/s²
Explanation:
To solve this problem let's start by finding the braking acceleration using kinematics, where the distance is x = 408 ft, the initial velocity vo = 100 mi / h and the final velocity is zero v = 0
v² = v₀² - 2 a x
0 = v₀² - 2ax
a =
Let's start by reducing the magnitudes to ft / s
v₀ = 100 mi / h (5280 foot / 1 mile) (1h / 3600 s) = 146.666 ft / s
let's calculate
a =
a = 26.36 ft / s²
Let's call this acceleration a_effective, this acceleration is in the opposite direction to the speed of the vehicle.
Let's use a rule of three (direct proportions) to find the acceleration applied by the brake system (a1) which has an efficiency of 95%. or 0.95
a₁ =
Let's use another direct proportion rule If the acceleration of the brake system (a₁) for an applied acceleration (a) with an efficiency of 0.90
a =
we substitute
a =
let's calculate
a =
a = 30.832 ft/s²
This is the maximum relationship that the vehicle can have for when it brakes to stop at the given distance