Question

A block of mass m = 2 kg slides back and forth on a frictionless horizontal track. It is attached to a spring with a relaxed length of L = 3 m and a spring constant k = 8 N/m. The spring is initially vertical, which is its the relaxed postion but then the block is pulled d = 3 m to one side

1. By what length is the spring extended? _______m

2. What is the potential energy stored in the spring? _____ j

3. The block is released. What is the maximum speed it attains? ______m/sec

4. Let's change the problem a bit. When the spring is vertical (hence, unstretched), the block is given an initial speed equal to 1 times the speed found in part (c). How far from the initial point does the block go along the floor before stopping? ____m

5.What is the magnitude of the acceleration of the block at this point (when the spring is stretched farthest)? m/sec^2

Answer 1

1) as the spring is vertical ....

so ... think of it as a triangle.. with hypotenuse as the final length .. and base as d = 3m and altitude = unstretched length = 3

so final length of spring = sqrt ( 3^2 + 3^2 ) = sqrt ( 18 ) = 4.2425 meters

so extension in spring = 4.2425 - 3 = 1.2425 metres

2) potential energy stored = 0.5 * k * x^2 = 0.5 * 8 * 1.2426^2 = 6.176 Joules

3) Let the macimum speed be v ...

maximum speed will be obtained when the block again reaches its initial position i.e. when the spring is fully untretched...

so at maximum speed... potential energy of spring = 0

so energy of system at maximum speed = 0.5 * m*v^2

inital energy of system = potential energy of spring = 6.176 J

as no friction is there... so energy can be conseved...

so intial energy = final ..

so 0.5 * 2 * v^2 = 6.176

so maximum velocty v = 2.49 m/sec

4) ..The velocity is the same

velocity = 1 * 2.49 = 2.49 m/sec

The distance won't change, it would be 3m.

5)

cos theeta = 3 / 4.2425

acceleration = k*x * cos theeta / mass = ( 8 * 1.2425* 3/ 4.2425 ) / 2 = 3.514 m/sec2