Physics hw helpppppp!!!?
A spring (oriented horizontally, k = 40 N/m) is attached to the left wall in a room with a frictionless floor. A mass (m = 0.67 kg), at rest on the floor, is attached to the right side of the spring. The spring length is 6 cm. Something begins to pull on the system to the right with an applied force of 35 N, until the spring is elongated by 4 cm and the force instantaneously disappears. What is the speed of the block when the applied force vanishes? What was the minimum spring length of the system?
- Anonymous8 months agoFavourite answer
F = k*x = 40*4/100 = 1.6 N
acceleration a = F/m = 1.6/(2/3) = 1.6*3/2 = 2.4 m/sec^2
V = 0
E = k/2*x^2 = 20*+4^2/10,000 = 320/10,000 = 0.032 joule
Lmin = 6-x = 2.0 cm
- AshLv 78 months ago
At 4 cm elongation, the spring Potential Energy = ½kx²
When the force disappears the spring PE is converted to Kinetic energy
½kx² = ½mv²
v² = kx²/m
v = x√(k/m)
v = 0.04√(40/0.67)
v = 0.3 m/s
speed of the block is 0.3 m/s
Without friction, all the Kinetic energy will get converted back to spring Potential energy and the spring will be compressed by 4cm.
So minimum spring length will be 6cm - 4cm = 2 cm