PHY1505 Assignment 4
(COMPLETE ANSWERS)
2024 - DUE September
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Complete, trusted
solutions and
explanations.
,1. A ball drops some distance and loses 30 J of gravitational potential energy.
Do NOT ignore air resistance. How much kinetic energy did the ball gain?
If the ball loses 30 J of gravitational potential energy, not all of this energy is converted into
kinetic energy due to air resistance. Air resistance dissipates some of the energy as heat and
sound, meaning the kinetic energy gained by the ball is less than 30 J.
However, without additional information (e.g., the amount of energy lost due to air resistance),
we can't determine the exact kinetic energy gained. It is simply less than 30 J.
2. The plot in the figure shows the potential energy of a particle, due to the
force exerted on it by another particle, as a function of distance. At which of
the three points labeled in the figure is the magnitude of the force on the
particle greatest?
To answer this question, recall that the force acting on a particle is related to the potential energy
function U(x)U(x)U(x) by the following relation:
F(x)=-du(x)/dx
This means that the force is proportional to the slope (or the steepness) of the potential energy
curve at a given point. The steeper the slope of the potential energy graph, the greater the
magnitude of the force.
If the slope is steep (either positive or negative), the force is large.
If the slope is flat, the force is zero.
So, to find where the force is greatest, you should identify the point with the steepest slope on the
potential energy graph. Without seeing the actual figure, the general rule is that the point where
the potential energy changes most rapidly (steepest slope) corresponds to the greatest magnitude
of force.
3. An 8.0-kg block is released from rest, with v1 = 0.00 m/s, on a rough
incline, as shown in the figure. The block moves a distance of 1.6-m down the
incline, in a time interval of 0.80 s, and acquires a velocity of v2 = 4.0 m/s.
How much work does gravity do on the block during this process?
,To calculate the work done by gravity on the block, we can use the formula for the work done by
a force:
Wgravity=Fgravity⋅d⋅cos(θ)W_{\text{gravity}} = F_{\text{gravity}} \cdot d \cdot
\cos(\theta)Wgravity=Fgravity⋅d⋅cos(θ)
where:
FgravityF_{\text{gravity}}Fgravity is the gravitational force component along the
incline,
ddd is the distance moved down the incline,
θ\thetaθ is the angle of the incline.
However, to simplify the calculation, we can also use the concept of energy. The work done by
gravity is equal to the change in the block's mechanical energy, which can be expressed as:
Wgravity=ΔKE+ΔPEfrictionW_{\text{gravity}} = \Delta KE + \Delta PE_{\text{friction}}
Wgravity=ΔKE+ΔPEfriction
Step 1: Calculate the change in kinetic energy (ΔKE\Delta KEΔKE):
ΔKE=12m(v22−v12)\Delta KE = \frac{1}{2} m (v_2^2 - v_1^2)ΔKE=21m(v22−v12)
Given:
m=8.0 kgm = 8.0 \, \text{kg}m=8.0kg
v1=0.00 m/sv_1 = 0.00 \, \text{m/s}v1=0.00m/s
v2=4.0 m/sv_2 = 4.0 \, \text{m/s}v2=4.0m/s
ΔKE=12(8.0 kg)((4.0 m/s)2−(0.00 m/s)2)\Delta KE = \frac{1}{2} (8.0 \, \text{kg}) \left( (4.0 \,
\text{m/s})^2 - (0.00 \, \text{m/s})^2 \right)ΔKE=21(8.0kg)((4.0m/s)2−(0.00m/s)2)
ΔKE=4.0 kg⋅(16.0 m2/s2)=64.0 J\Delta KE = 4.0 \, \text{kg} \cdot (16.0 \,
\text{m}^2/\text{s}^2) = 64.0 \, \text{J}ΔKE=4.0kg⋅(16.0m2/s2)=64.0J
Step 2: Calculate the gravitational potential energy change (ΔPE\Delta PEΔPE).
The change in potential energy depends on the height change, but we are asked for the work
done by gravity itself, which corresponds to the gain in kinetic energy in the absence of friction.
Therefore, the work done by gravity is related directly to the kinetic energy gained by the block.
Thus, the work done by gravity is:
Wgravity=64.0 JW_{\text{gravity}} = 64.0 \, \text{J}Wgravity=64.0J
This is the amount of work gravity does on the block during the process.
, 4. You do 174 J of work while pulling your sister back on a swing, whose
chain is 5.10 m long. You start with the swing hanging vertically and pull it
until the chain makes an angle of 32.0° with the vertical with your sister at
rest. What is your sister's mass, assuming negligible friction?
The work you do in pulling your sister on the swing is equal to the gravitational potential energy
gained by her as she is lifted higher. The formula for gravitational potential energy is:
PE=mghPE = mghPE=mgh
where:
PEPEPE is the potential energy,
mmm is the mass of your sister,
ggg is the acceleration due to gravity (9.81 m/s29.81 \, \text{m/s}^29.81m/s2),
hhh is the vertical height your sister is lifted.
Step 1: Find the vertical height hhh.
You pull the swing to an angle of 32.0∘32.0^\circ32.0∘ with the vertical, and the length of the
swing chain is 5.10 m. The vertical height hhh can be found by determining how much higher
your sister is relative to the lowest point of the swing. The height hhh is given by:
h=L−Lcos(θ)h = L - L \cos(\theta)h=L−Lcos(θ)
where:
L=5.10 mL = 5.10 \, \text{m}L=5.10m is the length of the swing,
θ=32.0∘\theta = 32.0^\circθ=32.0∘.
Substituting in the values:
h=5.10 m−5.10 m⋅cos(32.0∘)h = 5.10 \, \text{m} - 5.10 \, \text{m} \cdot
\cos(32.0^\circ)h=5.10m−5.10m⋅cos(32.0∘)
Step 2: Calculate hhh.
Using a calculator to find cos(32.0∘)\cos(32.0^\circ)cos(32.0∘):
