A mass is launched up an incline plane as shown. The spring has spring constant

Question

A mass is launched up an incline plane as shown. The spring has spring constant k. The plane is frictionless. The spring is compressed a distance and the mass m leaves the spring when the spring reaches its equilibrium position. How far does the mass slide along the plane before it stops and slides back down again?

3. (15 points A mass is launched up an incline plane as shown. The spring has spring constant k. The plane is frictionless. The spring is compressed a distance Ax and the mass m leaves the spring when the spring reaches its equilibrium position. How far does the mass slide along the plane before it stops and slides back down again? Ay

Explanation / Answer

We will solve this using energy conservation.

The spring has spring constant k. When the spring is compressed by distance x, the potential energy stored in the spring is PEspring = 1/2k(x)2 (we assume that spring has zero potential energy when compression is zero.)

When the mass compresses the spring, the speed of mass (when the compresion of spring is x) = zero which means the Kinetic energy (KE) of the mass is zero

So intital energy of the mass+spring system = (PE+KE)initial = 1/2k(x)2 (we assume the zero level of the gravitational potential energy of the mass to be when it has compressed the spring by x)

When the mass is released let us suppose it travels distance x on the inclined plane beyond the equilibrium position of the spring before it stops.

When the mass stops after traveling distance x, its kinetic energy is zero but it has acquired potential energy.

PE of the mass in this final position = mgh (where h = vertical height ascended by the mass after it was released by the spring with comrpession = x and travels along the plane by distance x+x )

So, h = (x+x) Sin

g = acceleration due to gravity

Now final energy of the mass+spring system = (PE+KE) = mgh = mg(x+x) Sin (spring is in relaxed state, so PE of spring = 0)

intital energy of the mass+spring system = final energy of the mass+spring system (as there is no dissipative force like friction)

or 1/2k(x)2 = mg(x+x) Sin

or mgx = 1/2k(x)2 - mgxSin

or x = [1/2k(x)2 - mgxSin]/mg

So, the mass slides distance 'x' along the plane before it stops and slides back down again.

You can ask for any more clarification or modification in the answer if you like...

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