Work Done on a System by an External Force — Question 1

Question 1

Applied work shared by motion and elastic deformation

Assume no heat transfer across the stated system boundary during the process unless otherwise specified. A 2 kg slider and its ideal horizontal spring form a system. The spring support is fixed. Initially the slider is at rest and the spring relaxed. An external agent does 25 J of work; final spring extension is 0.50 m and k=40 N/m. There is no friction.

1. Find the change in spring potential energy.

2. Find final kinetic energy and speed.

3. Why would setting the external work equal to kinetic-energy change alone fail?

Question 1 - Solution

Applied work shared by motion and elastic deformation

(a) Evaluate the elastic store.

The spring begins relaxed, so $$\boxed{\Delta U_s=\tfrac12kx_f^2=\tfrac12(40)(.50)^2=5\ \mathrm J}.$$

(b) Account for both system energy forms.

External work changes both system stores: $$25=\Delta K+5\Rightarrow\boxed{K_f=20\ \mathrm J},$$ $$\boxed{v_f=\sqrt{2K_f/m}=\sqrt{20}=4.472\ \mathrm{m/s}}.$$

(c) Distinguish external from net particle work.

For the chosen slider–spring system, $$W_{\mathrm{ext}}=\Delta K+\Delta U_s.$$ For the slider alone, the equivalent relation is $\Delta K=W_a+W_s=25-5$. Equating external input to kinetic gain alone would omit the 5 J stored in deformation.