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PHY1505 Assignment 4 (QUESTIONS & ANSWERS) 2024
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PHY1505 Assignment 4 Full Solutions 2024 ;100 % TRUSTED workings, Expert Solved, Explanations and Solutions. For assistance call or W.h.a.t.s.a.p.p us on ...(.+.2.5.4.7.7.9.5.4.0.1.3.2)........... 1. A ball drops some distance and loses 30 J of gravitational potential energy. Do NOT ignore air res...

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  • September 23, 2024
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PHY1505
ASSIGNMENT 4 2024
UNIQUE NO.
DUE DATE: 2024

, PHY1505

Assignment 4 2024

Unique Number:

Due Date: 2024

Mechanics (Physics)

1.

When the ball drops and loses 30 J of gravitational potential energy, not all of that
energy goes into kinetic energy because of air resistance. Some energy gets lost as
heat and sound due to friction with the air. So, the kinetic energy the ball gains will be
less than 30 J. Unfortunately, without knowing how much energy was lost to air
resistance, we can't determine the exact amount of kinetic energy gained, but we know
it’s less than 30 J.




2.

The magnitude of force exerted on a particle is related to how steep the slope of the
potential energy curve is. The steeper the slope, the greater the force. In the plot, you’d
look for the point where the potential energy curve is steepest, because that’s where the
force is largest. Without seeing the figure, this would typically occur where the curve
changes most rapidly.




3.

Gravity does work on the block by pulling it down the incline. The formula for work done
by gravity is related to the change in kinetic energy. Here, the block gains speed from 0
to 4.0 m/s over a distance of 1.6 m in 0.80 seconds.

, We use the work-energy principle:
W=ΔKE=12m(v22−v12)

W = \Delta KE = \frac{1}{2} m(v_2^2 - v_1^2)

W=ΔKE=21m(v22−v12)

Where v2=4.0 m/s,v1=0.00 m/s,m=8.0 kg

v_2 = 4.0 \, \text{m/s}, v_1 = 0.00 \, \text{m/s}, m = 8.0 \, \text{kg}

v2=4.0m/s,v1=0.00m/s,m=8.0kg.

So, W=12(8.0)(4.02−0.002)=12(8.0)(16)=64 J.

W=21(8.0)(4.02−0.002)=21(8.0)(16)=64J.

Gravity does 64 J of work on the block.

4. What is your sister’s mass if you do 174 J of work pulling her back on a swing?

The work you do pulling your sister back is stored as potential energy in the swing when
it’s at an angle of 32.0° from the vertical. This potential energy is given by:

W=mghW = mghW=mgh

Where W=174 J,g=9.8 m/s2, and hhh is the height the swing was lifted.

Using some trigonometry, the height hhh is h=L−Lcos⁡ (32.0∘), where L=5.10m is the
length of the swing’s chain. Plugging in the numbers:

h=5.10−5.10×cos⁡ (32∘)=5.10−4.32=0.78 m

h = 5.10 - 5.10 \times \cos(32^\circ) = 5.10 - 4.32 = 0.78 \, \text{m}

h=5.10−5.10×cos(32∘)=5.10−4.32=0.78m

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