Section 3: Work, Energy and Collisions

Question 1

If given L (the length of the cord in its relaxed state), 𝓶 (the mass of the jumper), the elastic constant K and the height (y) of the bridge from the jumper’s legs. How could one find distance (h) the jumper will be from the ground at the lowest point of the jump ?

Answer 1

We can start by setting up our Conservation of Energy Equation:

U₁ + K₁ = U₂ + K₂

We can eliminate K₁ since our initial velocity is 0, we can Convert K₂ into elastic energy since it is under elasticity after point L, we will then set the spring distance to (d) and solve for it first:

𝓶𝑔(y) = 𝓶𝑔(y - L - d) + ½Kd² → 𝓶𝑔(L + d) = ½Kd²

We can now solve for d using quadratic equation:

0 = ½Kd² - 𝓶𝑔d - 𝓶𝑔L

We can then use h = y - [L + d] to solve for the distance from the ground.

Question 2

A small block with mass 𝓶 slides without friction around the loop-the-loop apparatus shown in the diagram. It is released from rest from point A (where its height is 4R higher than the bottom of the circular section, which has a radius R. At point B, it is at the same height as the center of the circle.

Find Velocity at point B

Answer 2

We can start by stating our Conservation of Energy equation:

U₁ + K₁ = U₂ + K₂

At point A we are at the full height 4R and at point B we are at R and our initial K cancels due to lack of initial velocity:

𝓶𝑔(4R) = 𝓶𝑔(R) + ½𝓶𝓥² → 𝓥 = √(𝑔[6R])

Question 3

(A) What is the net force on Block A
(B) What is the acceleration of the system ?

Answer 3

(A) ΣFₐ = [𝓶ₐ𝑔 - T]

(B) 𝑔[𝓶ₐ - 𝓶𝔟]/[𝓶ₐ + 𝓶𝔟]

Question 4

For objects near the surface of the Earth, the universal law of gravitation can be simplified to F = mg, where g = 9.81m/s².

(A) If the mass of the Earth were doubled while at the same time its radius remained constant, by what factor would this change its acceleration due to gravity at it’s surface?

(B) If the radius of Earth doubles ?

Answer 4

(A) Since our formula is GmM/R², we know the acceleration due to gravity will double

(B) Given the same formula, we know the radius will cause the acceleration to lower by 1/4th

Question 5

What is the Equation to period of revolution (T) according to Keppler’s Law ?

Answer 5

T² = (4π²r³)/(GM)

Question 6

The satellite has mass m and the Earth has mass M and radius R. In order to be geosynchronous, the satellite must be at a certain height H above the Earth’s surface.

How do you solve for H ?

Answer 6

We start knowing the equation to Period of Revolution given our known variables and replacing our r variable with the R and H values:

T² = (4π²r³)/(GM) → T² = (4π²[R +H]³)/(GM)

Now we can manipulate this equation to get the value of H:

T² = (4π²[R +H]³)/(GM) → H = ³√( T²GM/4π² ) - R

Question 7

A box with a mass 𝓶 = 63kg is being pulled by a constant force F = 155N at an angle θ = 33°. The initial speed of the box is zero m/s.

What is an expression for the Work done on the Block ?

What is speed of block at d = 3.850m ?

Answer 7

(A) The work done can be found using Fcosθd

(B) Since we have 2 equations for work, we can set them equal and solve for Velocity:

Fcosθd = ½mv² → v = √(Fcosθd / ½m)

Question 8

What is the conversion from Rotational Velocity (ω) to Linear Velocity (𝓥) ?

Answer 8

𝓥 = rω

Question 9

What is equation of an angle (θ) relative to arch length (s) and radius (r) ?

Answer 9

θ = s/r

s = arch length
r = radius

Question 10

If a Bird can Distinguish θ = 3x10⁻⁴rad from 100m away

(A) How many degrees is this

(B) How long is this in arc length ?

Answer 10

(A) θ = 3x10⁻⁴rad (180/π) = 0.0172°

(B) Since we know θ = s/r → rθ = s

[100m][3x10⁻⁴rad] = 0.03m

Question 11

Two satellites orbit with radius of 4.23x10⁷m and 2.0° of seperation.

What is the arc length ?

Answer 11

First convert degrees to radians:

θ = 2.0° (π/180) = 0.0349rad

then use θ = s/r → rθ = s

[0.0349rad][4.23x10⁷m] = 1.48x10⁶m

Question 12

What is the equation to Tangent acceleration from Rotational Variables ?

Answer 12

a_t = rα

Question 13

What is the equation to Centripetal Acceleration from Rotational Variables ?

Answer 13

a_r = rω²

Question 14

If an object moves 1 revolution in 4.0s around a 1.2m center at constant velocity. What is the Linear Velocity and Linear acceleration ?

Answer 14

Velocity = rω

ω = 1.0rev/4.0s = 2π/4.0s = 1.6rev/s
𝓥 = rω = [1.6rev/s][1.2m] = 1.92m/s

Acceleration = a_r = rω²; a_t = rα
a_t = 0
a_r = [1.2m][1.6rev/s]² = 3.1m/s²

Question 15

What is equation to Torque (Շ) ?

Answer 15

Շ = r⟂F

where F is perpendicular to the moment of inertia.

Question 16

A block of mass m₁ = 18kg slides along a horizontal surface (with friction, μk = 0.39) a distance d = 2.85 m before striking a second block of mass m₂ = 6.25kg. The first block has an initial velocity of 𝓥 = 8.75m/s.

(A) How can we find the velocity of the second block after collision if the first is stopped ?

(B) How to find distance it took for m₂ to be stopped ?

Answer 16

Start by using our energy equation to find the velocity of the first block at the moment of the collision:

½m,𝓥₁² = ½m𝓥₁𝘧² + mgμd → v₁𝘧 = √(2[½𝓥₁² - gμd])

Now use this equation to solve the collision function for 𝓥₂:

m₁𝓥₁𝘧 = m₂𝓥₂ → 𝓥₂ = [m₁𝓥₁𝘧]/[m₂]

(B) Use the energy equation for m₂ and solve for d₂

½m₂𝓥₂² = m₂gμd₂ → d₂ = ½𝓥₂²/gμ

Question 17

How to find work if F(x) = a/x ?

Where a is a constant

Answer 17

Since Work = ∫F(x)dx

∫F(x)dx → a ∫ 1/x dx

W = a[ln(x)] for x = [b,c]

Question 18

A 60.0kg skier with an initial speed of 14m/s coasts up a 2.50m high rise as shown in the figure.

How to find velocity at top of hill ?

Answer 18

Start by finding the distance d of the hill using trig:

d = 2.5m/sin(35)

then set up the Energy Equation:

½m𝓥² = mgh + mgcos(θ)μd + ½m𝓥₂²

Then solve for 𝓥₂ using algebra:

𝓥₂ = {2/m]}

Question 19

What are the states of equilibrium for each point ?

Answer 19

A = Stable
B = Unstable
C = Neutral
D = Unstable
E = Stable
F = Indeterminate

Question 20

A bowling ball of mass m =1.1kg drops from a height h =11.7m. A semi-circular tube of radius r = 6.4m rests centered on a scale.
The scale reads in Newtons.

What is the equation of force at the bottom of the semicircle ?

Answer 20

Since at the bottom of the Semi Circle the system forces active will be both Fg & Fc, we can write our weight as:

W = mg +mv²/r

and since we know v², in a system which mass and gravity is active is v² = 2gh:

W = mg + 2ghm/r

Question 21

The force required to compress a non-standard spring as a function of displacement is given by the equation

F(x) = -Asin(bx) + kx,

where A = 11N, b = 11rad/m, and k = 68N/m.

How to solve for Work at any bounds of x?

Answer 21

Work can be solved for by simply taking the integral of F(x) over the given bounds:

W = ∫F(x)dx

W = ∫ -Asin(bx) + kx

W = Acos(bx)/x + ½Kx² | x = [c,d]

Note that the cos values are in radians

Question 22

A rotating platform of radius R = 3.71cm
is initially at rest. It then begins to move, such that a point on its edge experiences a constant tangential acceleration of a_t = 2.89cm/s²

What is the equation for Centripetal Acceleration in terms of rotational variables ?
.

Answer 22

We will start by identifying the Tangential Acceleration expression and solving for ω

a𝓉 = rα → a𝓉 = r(ω/t) → ω = ta𝓉/r

Now we can plug this into our Centripetal Acceleration expression: a𝖈 = rω²

a𝖈 = r[ta𝓉/r]² → a𝖈 = t² a𝓉² /r

Question 23

A 750kg van slows to rest from a speed of 27.08m/s in a distance of 125m. Assume the van is initially traveling in the positive direction.

(A): What is the force to stop the van in 125m ?

(B) What about 2m ?

Answer 23

(A) We can use the energy equation and solve for F:

½mv² = Fd. → F = ½mv² / d

F = ½[750kg][27.08m/s]² / -125m = -2200N

(B) We can use the same equation but replace 125m with 2m.

Question 24

What is the equation of rotational Kinetic Energy ?

Answer 24

KE = ½ɪω²

Where ɪ is based on the type of object.

Question 25

What is the equation of Torque for a large number of particles ?

Answer 25

mr²a

Question 26

Swimmer shown in the figure exerts an average horizontal backward force of F = 87.5N with his arm during each d = 1.80m long stroke.

(A) What is his work output, in joules, for each stroke?

(B) Calculate the power output, in watts, of his arms if he does 120 strokes per minute.

Answer 26

(A) The work can be found using W = Fd → [87.5N][1.80m] = 157.5J

(B) Power is by dividing work and time:

120st/min [1m/60s] = 2st/sec

Power = [157.5J][2st/s] = 315W

Question 27

On a winter day, you use a space heater to heat your bedroom for 30 minutes. If the power used by the space heater is 1000 W, then what is the energy consumed?

Answer 27

We can find energy by converting 30min into one hr then multiplying into our Power:

30min = .5hr

1000W * .5hr = 500W*h

E = .5kW*h