Question

Engineers have invented a new kind of spring whose restoring force is proportional to the third power of displacement: |F(x)| = |βx³| where B = (1/9) N/m³. One end of this spring is fixed to the bottom of an inclined plane which makes an angle θ = 36.87° with respect to the horizontal, and the other end is stretched up the incline and attached to a block of mass m = 3.00 kg. The spring is initially stretched a distance d = 6.00 m beyond its equilibrium length. The block is given an initial velocity vo = 5.00 m/s down the incline. Assume g = 10.0 m/s/s.

 (a) Calculate the initial acceleration of the block. Can we use this acceleration in the formula v² = v₀² + 2a_xΔx to find the speed after the block has slid a distance Δx? Explain your reasoning. 

 (b) The block has slid down the incline to the point at which the spring is no longer stretched and is back to its equilibrium length. Compute the work done on the block as it it slides through this displacement from (i) gravity, (ii) normal force, (iii) spring force.

 (c) What is the velocity of the block after it has slid down the incline to the point the spring is back to its equilibrium length?

Answer 1

We need to understand the forces acting on the mass at the top of the plane. We cannoty use the above equation right away since we have a two dimensional motion.

Let's begin our analysis.

It is acted by three forces, its weight (W = mg) ; the spring force (F = 1/9 x³) and the normal force.

We can simplify the analysis by setting the surface parallel to the plane as our x - axis. On this axis, we have the spring force and the x-component of the weight of the object.

To determine this component, we need to recall some angle relations. Then we can use trigonometric functions to determine the x component which is in this case is .

Recalling Newton's second law, Fnet = ma this becomes

then we have . Replacing with the given values,

  will yield which is more likely since the spring and gravity pull it down.

2) The work done on the block is the product of the force and the distance it moved along the direction of the force.

For gravity, work = change in its potential energy.

From the top of the inclined with to the equilibrium position. We can determine the change in height by using our sine function.

Since we know the angle and the hypotenuse, the opposite side can be determined and is equal to 3.6 m .

We can the use this height to find the work done

ii) The work done against the normal force is zero since the displacement is not parallel to the direction of the force.

iii) For the spring force the work done is

c) The veloctiy of the mass after coveing 6 m, will be known using the equation so

so .