# 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?

### 2 Answers

- 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

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- 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

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