Zabedul asked in Science & MathematicsPhysics · 1 year ago

m1 = 1.6 kg block slides on a frictionless horizontal surface and is connected on one side to a spring (k = 45 N/m) as shown in the figure above. The other side is connected to the block m2 = 3.2 kg that hangs vertically. The system starts from rest with the spring unextended.

a) What is the maximum extension of the spring? (*Note: I have tried 0.7m but it doesn't work)

b) What is the speed of block m2 when the extension is 65 cm?

Relevance
• Anonymous
1 year ago

a) y = max extension of spring

PE lost by m₂ = energy stored in spring

m₂gy = 0.5ky²

y = (m₂g)/(0.5k)

= (3.2 x 9.8)/(0.5 x 45)

= 1.4m

b) v = speed of block when x (extension) = 0.65

Energy conservation:

0.5kx² + 0.5(m₁ + m₂)v² = m₂gx

v² = (3.2 x 9.8 x 0.65 – 0.5 x 45 x 0.65²)/(0.5 x 4.8)

v = 2.1m/s.

• ?
Lv 7
1 year ago

The REASON 0.7 doesn't work is this is the distance at which the forces are in balance.  ie if you put it there it would stay there.  But this means it is the distance at which the system STOPS ACCELERATING. ie it is at its maximum speed at this distance.

In fact it will have gained kinetic energy and it will go EXACTLY twice this distance to come to a stop if there is no friction.

It will come to a stop when ALL of the energy has gone into the spring and nothing is moving.  ie 1/2 k x^2 = mgx

x = 2 mg/k  ie exactly twice the value of mg/k that you used.  Incidentally at this extension the force of the spring will also be exactly twice the force of gravity on the hanging mass so it will accelerate backwards and perform SHM. But I digress.

At 65 cm find the energy from gravity and the energy in the spring

1/2 (M+m) v^2 = mgh + 1/2 k h^2  Now put in the values and solve for v