A block of mass 0.40 kg slides in a straight line with a constant speed of 0.30 m s-1 along a horizontal surface, as shown in Fig. 3.1.

Question 11

(a) State what is meant by

(i) work done, [1]

(ii) elastic potential energy. [1]

(b) A block of mass 0.40 kg slides in a straight line with a constant speed of 0.30 m s^-1 along a horizontal surface, as shown in Fig. 3.1.

Fig. 3.1

The block hits a spring and decelerates. The speed of the block becomes zero when the

spring is compressed by 8.0 cm.

(i) Calculate the initial kinetic energy of the block. [2]

(ii) The variation of the compression x of the spring with the force F applied to the spring is shown in Fig. 3.2.

Fig. 3.2

Use your answer in (b)(i) to determine the maximum force FMAX exerted on the spring by

the block.

Explain your working. [3]

(iii) Calculate the maximum deceleration of the block. [1]

(iv) State and explain whether the block is in equilibrium

1. before it hits the spring,

2. when its speed becomes zero. [2]

(c) The energy E stored in a spring is given by

E = ½ k x²

where k is the spring constant of the spring and x is its compression.

The mass m of the block in (b) is now varied. The initial speed of the block remains constant and the spring continues to obey Hooke’s law.

On Fig. 3.3, sketch the variation of the maximum compression x₀ of the spring with mass m.

Fig. 3.3

[2]

[Total: 12]

Reference: Past Exam Paper – March 2016 Paper 22 Q3

Solution:

(a)

(i) Work is defined as the product of a force and the distance moved in the direction of the force.

(ii) Elastic potential energy is the energy stored (in an object) due to its extension / compression / change of shape.

(b)

(i)

E_K = ½ mv²

E_K = 0.5 × 0.40 × 0.30²

E_K = 1.8 × 10^–2 J

(ii)

{From the conservation of energy, the loss in KE is equal to the gain in elastic PE as the spring is compressed.}

(change in) kinetic energy = work done on spring / (change in) elastic potential energy     

{Elastic PE is given by the area under the compression-extension graph.}

1.8 × 10^–2 = ½ × F_MAX × 0.080

F_MAX = 0.45 N

(iii)

{F = ma}

a = F / m = 0.45 / 0.40 = 1.1 m s^–2

(iv)

1.

{For equilibrium, the resultant force should be zero.}

Since the block was moving at constant speed, its acceleration is zero. So, there is no resultant force and the block is in equilibrium.

2.

There is a deceleration, so the resultant force is not zero. The block is not in equilibrium.

(c)

curved line from the origin with decreasing gradient

{The stored energy is obtained from the kinetic energy of the block.

½ mv² = ½ kx²

This simplifies to x² = mv²/k

Since v and k are considered to be constant, x² ∝ m

So, the compression x ∝ √(m)

The graph of y = √x is a curve with decreasing gradient, starting at the origin.}