How much does the spring compress?

A 2.0 - kg mass is dropped 2.0 m above a spring with a spring constant 40.0 N/m. How much does the spring compress? Use g = 10 m/s².

Solution

Given Data:
m = 2.0kg
k = 40.0N/m
h = 2.0m
x = ?

Useful formulas:
Uₛ = ½kx²
Uₘ = mgh

Solution:
mg(h + x) = ½kx²
½kx² - mgx – mgh = 0

x² - (2mg/k)x – (2mg/k)h = 0

x² – 2 (mg/k)x + (mg/k)² = (mg/k)² + (2mg/k)h

(x - mg/k)² = (mg/k)² + (2mg/k)h

x - mg/k = ±√ {(mg/k)² + 2(mg/k)h}

x = mg/k±√ {(mg/k)² + 2(mg/k)h}

Calculation
mg/k = 2kg*10m/s ² / 40N/m = ½m

(mg/k)² = ¼m²

2(mg/k)h = 2·½m·2m = 2m²

(mg/k)² + 2(mg/k)h = ¼m² + 2m² = (⁹/₄)m²

√ {(mg/k)² + 2(mg/k)h} = (³/₂)m

x₁ = ½m + (³/₂)m = 2m
x₂ = - ½m - (³/₂)m = - 1m

Problem's answer: 2.0 m

Physics Problem - Potential Energy

The potential energy is given by the function U(x) = kx⁻¹, where k is a constant, and variable x is the position, the coordinate. (a) Derive the function of the coordinate x, which describes the force produced by this potential energy.

F(x) = - d U(x) / dx

F(x) = - d kx⁻¹ / dx = kx⁻²

F(x) = kx⁻²

(b) Let this potential energy is obtained by the body with the mass of 10kg. The constant is known: k = -3.98 × 10¹⁵ Jm.
Find the force on this body when it is located atx = 6.38 × 10⁶ m?

F(x) = kx⁻² = -3.98 × 10¹⁵ Jm / ( 6.38 × 10⁶ m)²

The code for the Google calculator: (-3.98E15J*m) / ( 6.38E6m )^2

Google calculator's result: -97.7781272 newtons

Problem's answer: F = -97.8 N or 97.8 N towards x = 0

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