How Much Work Is Done by the Spring on the Mass?

A horizontal spring having constant k is attached to a block having mass M. The mass is oscillating freely on a frictionless table with amplitude A. As the mass travels from the position of the maximum stretch to half the position of maximum stretch, how much work is done by the spring on the mass?

Given Data

k = 50.0 N/m

M= 0.50 kg

A= 15.0 cm

xₒ=A

x=A/2

W=?

Solution

U=½kx²

Uₒ=½kxₒ²

Uₒ-U=W

W=½kxₒ²-½kx²=½k(xₒ²-x²)

W=½k(A²-(½A)²)=½k(A²-(½A)²)=½k(A²-¼A²)=½k(¾A²)=⅜kA²

W=⅜·(50 N/m)·(0.15m)²

Code for the Google Calculator: 3/8*(50 N/m)*(0.15m)^2

Google Calculator's result: 0.421875 joules

Answer:   W=0.42J

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