Question

005 (part 2 of 3) 10.0 points
b) the floor?
answer in units of j.

006 (part 3 of 3) 10.0 points
c) a point at the same elevation as the ball?
answer in units of j.

007 10.0 points
a spring with a force constant of 5.2 n/m has
a relaxed length of 2.43 m. when a mass is
attached to the end of the spring and allowed
to come to rest, the vertical length of the
spring is 3.57 m.

answer in units of m/s.

014 (part 1 of 3) 10.0 points
a flight attendant pulls her 62.4 n flight bag
a distance of 287 m along a level airport floor
at a constant velocity. the force she exerts is
35.0 n at an angle of 63.5° above the horizon-
tal.
a) find the work she does on the flight bag.
answer in units of j.

015 (part 2 of 3) 10.0 points
b) find the work done by the force of friction
on the flight bag.

haynes (hlh2749) - energy 1 - neff - (76523) 2
answer in units of j.

016 (part 3 of 3) 10.0 points
c) find the coefficient of kinetic friction be-
tween the flight bag and the floor.

017 (part 1 of 2) 10.0 points
a horizontal force of 150 n is used to push a
50.0 kg packing crate a distance of 5.00 m on
a rough horizontal surface.
the acceleration of gravity is 9.81 m/s².
if the crate moves with constant velocity,
calculate
a) the work done by the force.
answer in units of j.
remains at rest.

1. it moves to the left of x₀ and does not

return.

1. it cannot reach either x₀ or x₂.

1. it oscillates with maximum position x₂

and minimum position x₀.

020 10.0 points
a small mass is released from rest at a very
great distance from a much larger stationary
mass.

Explanation:

Since there are multiple sub - questions, let's take the sub - question 014 part (a) as an example. The discipline is Natural Science, sub - field is Physics (specifically related to work and energy in mechanics). We will use the Step - by - Step Format.

Step 1: Recall the work formula

The work done by a force \(F\) over a displacement \(d\) is given by \(W = Fd\cos\theta\), where \(\theta\) is the angle between the force and the displacement. Here, the force exerted \(F = 35.0\space N\), the displacement \(d=287\space m\), and the angle \(\theta = 63.5^{\circ}\) (the angle between the applied force and the horizontal, and the displacement is horizontal, so we use this angle in the cosine term).

Step 2: Calculate the cosine of the angle

First, calculate \(\cos(63.5^{\circ})\). Using a calculator, \(\cos(63.5^{\circ})\approx0.4493\)

Step 3: Substitute values into the work formula

Substitute \(F = 35.0\space N\), \(d = 287\space m\) and \(\cos\theta=0.4493\) into \(W=Fd\cos\theta\).
\(W=(35.0\space N)\times(287\space m)\times0.4493\)
First, calculate \(35.0\times287 = 10045\)
Then, \(10045\times0.4493\approx10045\times0.45 = 4520.25\) (a more accurate calculation: \(10045\times0.4493=10045\times(0.4 + 0.04+0.0093)=10045\times0.4+10045\times0.04 + 10045\times0.0093=4018+401.8+93.4185 = 4513.2185\space J\))

Answer:

\(\approx4510\space J\) (or more precisely \(4513\space J\) depending on the calculator precision)