AP Physics C: Mechanics:: Work Energy Power

What is the GENERAL, calculus-based DEFINITION of WORK done by a VARIABLE force F(x) as an object moves along the x-axis from position x1 to position x2, using a DEFINITE INTEGRAL?

- **A)** W = F(x2) - F(x1) (subtracting force VALUES rather than integrating), rather than the correct definite-integral definition
- **B)** W = INTEGRAL from x1 to x2 of F(x) dx -- this definite integral definition correctly calculates work for ANY force function, including forces that VARY with position, extending far beyond the simple W=F*d formula (valid only for constant force)
- **C)** Work can only be calculated using integration when force is CONSTANT, with no actual applicability to variable, position-dependent forces
- **D)** This concept has no actual mathematical relationship between work and the definite integral of a position-dependent force function

For a CONSTANT force F acting over a displacement from x1 to x2, how does the GENERAL calculus-based work formula, W=INTEGRAL[F(x)]dx, correctly REDUCE to the familiar algebra-based formula, W=F*d (where d=x2-x1)?

- **A)** When F(x)=F is CONSTANT, the integral simplifies directly: W = INTEGRAL from x1 to x2 of F dx = F*(x2-x1) = F*d -- exactly recovering the simpler algebra-based formula as a SPECIAL CASE of the more general integral definition
- **B)** The integral definition of work can ONLY be used for variable forces, with the constant-force case requiring an entirely separate, unrelated formula
- **C)** A constant force produces a COMPLETELY DIFFERENT work value when using the integral definition compared to the simple W=Fd formula, contradicting their actual special-case relationship
- **D)** The general calculus-based work formula has no actual relationship to the simpler algebra-based W=Fd formula, even for a constant force

For a SPRING FORCE, F(x)=-kx (following Hooke's Law), what is the WORK DONE BY THE SPRING as it moves from x=0 (natural length) to x=x1 (some displacement), calculated using the definite integral W=INTEGRAL[F(x)]dx?

- **A)** W = INTEGRAL from 0 to x1 of (-kx) dx = [-(1/2)*k*x^2] from 0 to x1 = -(1/2)*k*x1^2 -- this NEGATIVE result reflects that the spring force OPPOSES the displacement (as x increases, the restoring force acts in the negative direction), and its MAGNITUDE matches the elastic potential energy formula from earlier physics courses
- **B)** W = -k*x1 (treating the integral as if it were a simple, direct evaluation of F(x) at x1, rather than integrating), rather than the correct integral result
- **C)** This concept has no actual mathematical relationship between integrating a spring's variable force and the resulting work calculation
- **D)** W = -(1/2)*k*x1 (incorrectly omitting the square on x1 from the correct integration result)

What is the general RELATIONSHIP between a CONSERVATIVE FORCE, F(x), and its associated POTENTIAL ENERGY function, U(x), expressed as a DERIVATIVE (specifically, F(x) as the NEGATIVE derivative of U(x))?

- **A)** F(x) = -dU/dx -- the force associated with a conservative potential energy function is the NEGATIVE of the derivative of that potential energy with respect to position; this relationship directly connects force and potential energy via calculus, complementing the earlier integral relationship (U(x) as the negative integral of F(x))
- **B)** F(x) = dU/dx (WITHOUT the negative sign), rather than the correct negative-derivative relationship
- **C)** This concept has no actual mathematical relationship between a conservative force and the derivative of its associated potential energy function
- **D)** F(x) = U(x) (force and potential energy being IDENTICAL functions), rather than being related via a derivative

For a spring's potential energy function, U(x)=(1/2)*k*x^2, what FORCE function F(x) is recovered by taking the NEGATIVE DERIVATIVE, F(x)=-dU/dx, and how does this CONFIRM the earlier work-integral derivation of the spring force?

- **A)** This concept has no actual relationship between differentiating a spring's potential energy function and recovering its associated force function
- **B)** F(x) = -dU/dx = -d/dx[(1/2)*k*x^2] = -kx -- this exactly RECOVERS Hooke's Law (F=-kx), confirming the CONSISTENCY between the earlier integral-based derivation (force to potential energy) and this derivative-based approach (potential energy back to force)
- **C)** F(x) = kx^2 (a completely different function from the expected Hooke's Law result), contradicting the actual consistency between the integral and derivative approaches
- **D)** F(x) = (1/2)*k*x (omitting the correct differentiation of the squared term), rather than the correct F=-kx result

What is the WORK-ENERGY THEOREM, expressed using calculus, showing that the WORK done by the NET FORCE equals the CHANGE IN KINETIC ENERGY, derived by INTEGRATING Newton's second law, m*(dv/dt)=F_net, with respect to POSITION (using the chain rule substitution dv/dt = v*dv/dx)?

- **A)** This calculus-based derivation only applies to CONSTANT forces, with no actual ability to derive the work-energy theorem for variable forces
- **B)** Starting from m*v*(dv/dx) = F_net(x) (using the chain-rule substitution), INTEGRATING both sides with respect to x gives: INTEGRAL[m*v*dv] = INTEGRAL[F_net(x)]dx, which evaluates to (1/2)*m*v2^2 - (1/2)*m*v1^2 = W_net -- exactly the work-energy theorem, DERIVED rigorously from Newton's second law using calculus
- **C)** This derivation shows that work equals the change in MOMENTUM (rather than kinetic energy), contradicting the actual work-energy theorem
- **D)** The work-energy theorem has no actual calculus-based derivation from Newton's second law using the chain-rule substitution

What is POWER, expressed using calculus, as the INSTANTANEOUS RATE at which WORK is done, in terms of the DERIVATIVE of the work function with respect to time?

- **A)** Power is defined as the SECOND derivative of work with respect to time, rather than the first derivative
- **B)** P(t) = dW/dt -- instantaneous power is the DERIVATIVE of work with respect to time, directly paralleling the earlier calculus-based definitions of velocity (dx/dt) and acceleration (dv/dt) as derivatives of other physical quantities
- **C)** P(t) = INTEGRAL[W]dt (treating power as the integral, rather than the derivative, of work), rather than the correct derivative-based definition
- **D)** This concept has no actual mathematical relationship between power and the derivative of work with respect to time

What is the mathematical RELATIONSHIP between INSTANTANEOUS POWER and the FORCE and VELOCITY of an object, derived by combining P=dW/dt with the definition of work (dW=F*dx) and velocity (v=dx/dt)?

- **A)** This concept has no actual mathematical relationship between instantaneous power, force, and velocity
- **B)** P = F/v (dividing force by velocity), rather than the correct P=Fv multiplicative relationship
- **C)** P = F*v (force multiplied by velocity) -- this follows directly from P=dW/dt=d(F*x)/dt=F*(dx/dt)=F*v (for a force acting parallel to velocity), providing a calculus-based derivation of this formula, which directly parallels the analogous formula from algebra-based physics
- **D)** Power depends only on force, with velocity playing no actual role in the P=Fv relationship