Analyze Energy Using Conservation Laws - Physics

  1. Identify the correct energy equation for a pendulum bob between two heights when air resistance is negligible.
    mghi+12mvi2=mghf+12mvf2mgh_i+\frac{1}{2}mv_i^2=mgh_f+\frac{1}{2}mv_f^2. Conservation of mechanical energy for pendulum motion.
  2. Find the speed vv of a mass mm after descending height hh with kinetic friction doing work −fkd-f_k d.
    v=2gh−2fkdmv=\sqrt{2gh-\frac{2f_k d}{m}}. Apply energy conservation with friction work included.
  3. What is the gravitational potential energy formula near Earth for mass mm at height hh?
    Ug=mghU_g=mgh. Weight (mgmg) times height gives gravitational potential energy.
  4. State the conservation of mechanical energy equation when only conservative forces act.
    Ki+Ui=Kf+UfK_i+U_i=K_f+U_f. Total mechanical energy remains constant with only conservative forces.
  5. Find the speed vv after dropping from rest through height hh with no losses.
    v=2ghv=\sqrt{2gh}. From mgh=12mv2mgh = \frac{1}{2}mv^2 with energy conservation.
  6. Identify the correct energy equation for a block starting from rest: slide down height hh with friction work WfW_f.
    mgh+Wf=12mv2mgh+W_f=\frac{1}{2}mv^2. Initial PE plus friction work equals final KE.
  7. What is the general energy accounting equation including nonconservative work?
    Ki+Ui+Wnc=Kf+UfK_i+U_i+W_{nc}=K_f+U_f. Includes nonconservative work in energy conservation.
  8. What is the general energy accounting equation including nonconservative work WncW_{\text{nc}}?
    Ki+Ui+Wnc=Kf+UfK_i+U_i+W_{\text{nc}}=K_f+U_f. Nonconservative work accounts for energy not conserved.
  9. What is the relationship between nonconservative work and mechanical energy change?
    Wnc=ΔEmechW_{nc}=\Delta E_{mech}. Nonconservative work equals the change in total mechanical energy.
  10. Find the speed at the bottom after dropping from rest through height hh, ignoring air resistance.
    v=2ghv=\sqrt{2gh}. From mgh=12mv2mgh=\frac{1}{2}mv^2, solving for vv gives 2gh\sqrt{2gh}.
  11. Find the speed vv at height hh for an object launched upward with speed v0v_0 (no air drag).
    v=v02−2ghv=\sqrt{v_0^2-2gh}. Energy conservation: 12mv02=12mv2+mgh\frac{1}{2}mv_0^2=\frac{1}{2}mv^2+mgh.
  12. Find the spring compression xx needed to stop a mass mm moving at speed vv on a frictionless surface.
    x=vmkx=v\sqrt{\frac{m}{k}}. All kinetic energy converts to spring PE: 12mv2=12kx2\frac{1}{2}mv^2=\frac{1}{2}kx^2.
  13. What is the work–energy theorem relating net work and kinetic energy change?
    Wnet=ΔKW_{net}=\Delta K. Net work equals the change in kinetic energy.
  14. State the work–energy theorem relating net work and kinetic energy change.
    Wnet=ΔKW_{\text{net}}=\Delta K. Net work equals the change in kinetic energy.
  15. What is the kinetic friction magnitude formula using coefficient μk\mu_k and normal force NN?
    fk=μkNf_k=\mu_k N. Kinetic friction equals coefficient times normal force.
  16. Find the minimum initial speed v0v_0 needed to reach height hh if nonconservative work is Wnc<0W_{nc}<0.
    v0=2gh−2Wncmv_0=\sqrt{2gh-\frac{2W_{nc}}{m}}. Rearrange energy equation with negative nonconservative work.
  17. Find the final speed vfv_f if net work on a mass mm is WnetW_{net} and initial speed is viv_i.
    vf=vi2+2Wnetmv_f=\sqrt{v_i^2+\frac{2W_{net}}{m}}. From work-energy theorem: Wnet=12m(vf2−vi2)W_{net} = \frac{1}{2}m(v_f^2 - v_i^2).
  18. Find the work done by friction WfW_f if mechanical energy decreases from EiE_i to EfE_f.
    Wf=Ef−EiW_f=E_f-E_i. Friction work equals the mechanical energy loss.
  19. Find the maximum height hh reached by an object launched upward with initial speed v0v_0 (no air drag).
    h=v022gh=\frac{v_0^2}{2g}. From 12mv02=mgh\frac{1}{2}mv_0^2=mgh, solving for hh gives v022g\frac{v_0^2}{2g}.
  20. What conservation equation applies when only conservative forces do work on a system?
    Ki+Ui=Kf+UfK_i+U_i=K_f+U_f. Mechanical energy is conserved when only conservative forces act.
  21. Identify the correct expression for mechanical energy lost to friction over distance dd on level ground.
    ΔEmech=−μkNd\Delta E_{mech}=-\mu_k N d. Friction force times distance gives energy lost.
  22. What is the elastic potential energy formula for a spring with constant kk stretched by xx?
    Us=12kx2U_s=\frac{1}{2}kx^2. Spring energy equals half the spring constant times displacement squared.
  23. What is the definition of total mechanical energy EmechE_{mech} in a system?
    Emech=K+UE_{mech}=K+U. Total mechanical energy is the sum of kinetic and potential energies.
  24. Identify the sign of ΔUg\Delta U_g when an object moves downward by Δh<0\Delta h<0.
    ΔUg<0\Delta U_g<0. Moving down means Δh<0\Delta h<0, so mgΔh<0mg\Delta h<0.
  25. Find the height hh reached if an object launched upward with speed v0v_0 stops at the top (no losses).
    h=v022gh=\frac{v_0^2}{2g}. From 12mv02=mgh\frac{1}{2}mv_0^2 = mgh at maximum height.
  26. What is the work done by a constant force FF over displacement dd at angle θ\theta?
    W=Fdcos⁡θW=Fd\cos\theta. Work equals force times displacement times cosine of angle between them.
  27. Find the speed vv of mass mm after sliding down height hh with kinetic friction μk\mu_k over distance dd.
    v=2g(h−μkd)v=\sqrt{2g(h-\mu_k d)}. Initial PE minus friction work equals final KE.
  28. Choose the correct statement: with only conservative forces, mechanical energy is (constant) or (decreases).
    Mechanical energy is constant. Conservative forces don't dissipate energy.
  29. What is the kinetic energy formula for a mass mm moving at speed vv?
    K=12mv2K=\frac{1}{2}mv^2. Energy of motion equals half the mass times velocity squared.
  30. What is the gravitational potential energy change when an object drops by height hh?
    ΔUg=−mgh\Delta U_g=-mgh. Negative because potential energy decreases when falling.
  31. Identify the correct condition for using mghmgh for gravitational potential energy.
    Near Earth with approximately constant gg. mghmgh assumes constant gravitational field strength.
  32. Find the minimum initial speed v0v_0 to reach height hh if a constant friction force fkf_k acts over distance dd.
    v0=2gh+2fkdmv_0=\sqrt{2gh+\frac{2f_k d}{m}}. Initial KE must overcome both PE gain and friction work.
  33. What is the gravitational potential energy change near Earth for height change Δh\Delta h?
    ΔUg=mgΔh\Delta U_g=mg\Delta h. Weight mgmg times height change gives gravitational PE change.
  34. Find the final speed vv if net work done on mass mm is WnetW_{\text{net}} and initial speed is viv_i.
    v=vi2+2Wnetmv=\sqrt{v_i^2+\frac{2W_{\text{net}}}{m}}. From work-energy theorem: Wnet=12m(v2−vi2)W_{\text{net}}=\frac{1}{2}m(v^2-v_i^2).
  35. Find the compression xx of a spring that stops a mass mm moving at speed vv on a frictionless surface.
    x=vmkx=v\sqrt{\frac{m}{k}}. From 12mv2=12kx2\frac{1}{2}mv^2 = \frac{1}{2}kx^2 by energy conservation.
  36. Which option is correct for a conservative force: Wc=−ΔUW_c=-\Delta U or Wc=+ΔUW_c=+\Delta U?
    Wc=−ΔUW_c=-\Delta U. Conservative force work equals negative of potential energy change.
  37. What is the sign of work done by kinetic friction on a moving object over distance dd?
    Wf=−fkdW_f=-f_k d. Friction opposes motion, so work is negative.
  38. What is the elastic potential energy stored in a spring with constant kk stretched by xx?
    Us=12kx2U_s=\frac{1}{2}kx^2. Spring PE equals half the spring constant times displacement squared.
  39. What is the work done by kinetic friction with coefficient μk\mu_k over distance dd on level ground?
    Wf=−μkmgdW_f=-\mu_kmgd. Friction opposes motion, so work is negative.

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