Answer:
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- <u>1. The potential energy of the swing is the greatest at the position B.</u>
- <u>2. As the swing moves from point B to point A, the kinetic energy is increasing.</u>
Explanation:
Even though the syntax of the text is not completely clear, likely because it accompanies a drawing that is not included, it results clear that the posittion A is where the seat is at the lowest position, and the position B is upper.
The gravitational <em>potential energy </em>is directly proportional to the height of the objects with respect to some reference altitude. Thus, when the seat is at the position A the swing has the smallest potential energy and when the seat is at the <em>position B the swing has the greatest potential energy.</em>
Regarding the forms of energy, as the swing moves from point B to point A, it is going downward, gaining kinetic energy (speed) at the expense of the potential energy (losing altitude). When the seat passes by the position A, the kinetic energy is maximum and the potential energy is miminum. Then the seat starts to gain altitude again, losing the kinetic energy and gaining potential energy, up to it gets to the other end,
Answer:
F = 69.3 N
Explanation:
For this exercise we use Newton's second law, remembering that the static friction force increases up to a maximum value given by
fr = μ N
We define a reference system parallel to the floor
block B ( lower)
Y axis
N - W₁-W₂ = 0
N = W₂ + W₂
N = (M + m) g
X axis
F -fr = M a
for block A (upper)
X axis
fr = m a (2)
so that the blocks do not slide, the acceleration in both must be the same.
Let's solve the system by adding the two equations
F = (M + m) a (3)
a =
the friction force has the formula
fr = μ N
fr = μ (M + m) g
let's calculate
fr = 0.34 (2.0 + 0.250) 9.8
fr = 7.7 N
we substitute in equation 2
fr = m a
a = fr / m
a = 7.7 / 0.250
a = 30.8 m / s²
we substitute in equation 3
F = (2.0 + 0.250) 30.8
F = 69.3 N
Weight equals mass times gravitational acceleration=400N, so mass=400/9.8=41kg approx.
The car would go from zero to 58.0 mph in 2.6 sec.
Since the force on the car is constant, therefore the acceleration of the car would also be constant.
Now for constant acceleration we can use the equation of motion
Using first equation of motion to calculate the acceleration of the car
v=u+at
29=0+a×1.30 ...... Eq. (1)
Again using the first equation of motion
58=0+a*t ....... Eq. (2)
Dividing eq. (2) with equation 1
t=2×1.3
t=2.6 sec
