A train with the mass M accelerates uniformly from the rest to the speed V when passing through a distance d

Problem from Quiz 5

A train with the mass M accelerates uniformly from the rest to the speed V when passing through a distance d.
What is its kinetic energy
when it passing through a distance d₁ (d₁ < d)?

Given Data: M = 15,000 kg

Vₒ = 0

V = 15 m / s

d = 150 m

d₁ = 75 m

K₁ = ?

Solution:

K(kinetic energy) = Kₒ(kinetic energy initial) + W(work) = W

W = Fx

K = W = Fd = ½MV²

F = ½MV² / d

K₁ = W₁ = Fd₁ = ½MV² d₁ / d

K₁ = ½(15,000 kg)(15 m/s)² (75m) / (150m) = 843750 J

Google Calculator Code:

0.5*(15000 kg)*(15 m/s)^2*(75m) / (150m)

Answer: 840 kJ

Physics Problem - Average Power

7.36. The electric motor of a 2-kg train accelerates the train from rest to 1 m/s in 1 s. Find the average power delivered to the train during the acceleration.

Solution

Given Data	Useful formulas
M=2kg V₀=0 V=1m/s t=1s pₐ=?	K=½MV² K₀=0 pₐ=W/t W= K-K₀	pₐ=W/t=(K-K₀)/t=½MV²/t pₐ = ½MV²/t pₐ= ½ (2kg) (1m/s)² /1s = =1 kg m/s²∙m/s=1N∙m/s=1J/s=1W pₐ=1W

7.40. A 1000-kg elevator starts from rest.
It moves upward for 4 s with constant acceleration until it reaches its cruising speed of 2 m/s.

(a) What is the average power of the elevator motor during this period?

(b) What is the motor power when the elevator moves at its cruising speed?

Given Data	Useful formulas
M=1000kg (a) V₀=0 V=2m/s t=4s pₐ=? (b) V=const=2m/s	K=½MV² K₀=0 U=Mgh pₐ=W/t W= =(K+U)-(K₀+ U₀) h=V₀t+at²/2 a=(V-V₀)/t p = F V =  = MgV	W= (K+U)-(K₀+ U₀)= K+U= =½MV²+Mgh=½MV²+Mgh h=at²/2=(V/t)∙t²/2=Vt/2 W= ½MV²+MgVt/2 pₐ=W/t=½MV²/t+MgV/2 or pₐ=W/t=M(V/t+g)∙½V ½MV²/t=½(1000kg)(2m/s)²/4s=500W MgV/2=1000kg∙10m/s²∙2m/s /2=10000W pₐ=10500W=10.5kW or pₐ=1000kg∙(2 m/s / 4 s +10 m/s²)∙½∙2 m/s= =10.5kW p=MgV=1000kg∙10m/s²∙2m/s=20000W=20kW

Deadline

The announcement for CityTech students:

The deadline for online quizzes 2,3,4,5, and 6 is October 6 (Thursday), 4:00 PM.

After the deadline, these quizzes will be closed for records, and correct answers will be displayed.

A car moving in the x direction has acceleration aₓ, that varies with time as shown in the figure

A car moving in the x direction has acceleration aₓ, that varies with time as shown in the figure. pic.twitter.com/xLnJUtQIyp

— Physics I (@PHY210) September 28, 2016

Problem from Online Quiz 1

The measured length of a cylindrical laser beam is 12.1 meters, its measured diameter is 0.00121 meters, and its measured intensity is 1.21 × 10⁵ W/m² . Which of these measurements is the most precise?

Solution:
Uncertainty of the measurement result of 12.1 m is 0.05 m so the precision can be shown by the relative error of 0.05m / 12.1m = 0.05/12.1

Uncertainty of the measurement result of 0.00121 m is 0.000005 m so the precision can be shown by the relative error of 0.000005m / 0.00121m = 0.05/12.1

Uncertainty of the measurement result of 1.21 × 10⁵ W/m² is 0.05 × 10⁵ W/m² so the precision can be shown by the relative error of (1.21 × 10⁵ W/m² ) / (0.05 × 10⁵ W/m²) = 0.05/12.1

As relative errors are the same then precision levels are the same.

PHY 215 Quiz 1 (eztestonline.com/695230/14274823527240100.tp4) Is Closed for Records. Nothing Recorded

PHY 215 Quiz 1 (eztestonline.com/695230/14274823527240100.tp4) is closed for records. Nothing recorded.

If a (t) is acceleration and a (t) = a₀ + bt, where a₀ and b are constants, and t is the time

If a (t) is acceleration and a (t) = a₀ + bt, where a₀ and b are constants, and t is the time.

v (t) = v₀ + a₀t + bt² / 2

x (t) = x₀ + v₀t + a₀t² / 2 + bt³ / 6

x (t₁) = x₀ + v₀t₁ + a₀t₁² / 2 + bt₁³ / 6

x (t₂) = x₀ + v₀t₂ + a₀t₂² / 2 + bt₂³ / 6

x (t₂) - x (t₁) = v₀ (t₂ - t₁) + a₀ (t₂² - t₁²) / 2 + b (t₂³ - t₁³) / 6