Resistive Forces
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AP Physics C: Mechanics › Resistive Forces
An object with mass $$m$$ is projected horizontally with initial velocity $$v_0$$ on a surface where the only horizontal force is a resistive force given by $$F = -kv$$. The velocity of the object as a function of time is $$v(t) = v_0 e^{-kt/m}$$. What is the total distance the object travels as it slows to a stop?
$$v_0 m / k$$
$$v_0 k / m$$
The object never stops, so the distance is infinite.
$$v_0 m / (2k)$$
Explanation
The total distance is the integral of the velocity function from $$t=0$$ to $$t=\infty$$. $$x_{total} = \int_0^\infty v(t) dt = \int_0^\infty v_0 e^{-kt/m} dt$$. The integral evaluates to $$v_0 [-\frac{m}{k}e^{-kt/m}]_0^\infty = -\frac{mv_0}{k}(0 - 1) = \frac{mv_0}{k}$$.
A boat of mass $$m$$ starts from rest at $$t=0$$. A constant force $$F_{app}$$ is applied, and it experiences a resistive force $$F_r = -kv$$. The boat's speed $$v$$ as a function of time $$t$$ is given by which expression?
$$v(t) = \frac{F_{app}}{m}(1 - e^{-kt/m})$$
$$v(t) = \frac{F_{app}}{m}t - \frac{k}{2m}t^2$$
$$v(t) = \frac{F_{app}}{k}(1 - e^{-kt/m})$$
$$v(t) = \frac{F_{app}}{k} e^{-kt/m}$$
Explanation
The differential equation for the boat's motion is $$m \frac{dv}{dt} = F_{app} - kv$$. This equation is mathematically analogous to that of a falling object, with the constant gravitational force $$mg$$ replaced by the constant applied force $$F_{app}$$. The solution, with the initial condition $$v(0)=0$$, is therefore $$v(t) = \frac{F_{app}}{k}(1 - e^{-kt/m})$$, where $$F_{app}/k$$ is the terminal velocity.
A particle of mass $$m$$ moves in a straight line with velocity $$v(t) = v_0 e^{-kt/m}$$. Its position at $$t=0$$ is $$x=0$$. What is its position $$x$$ as a function of time $$t$$?
$$x(t) = v_0 t - \frac{k}{2m}t^2$$
$$x(t) = -\frac{mv_0}{k} e^{-kt/m}$$
$$x(t) = v_0(1 - e^{-kt/m})$$
$$x(t) = \frac{mv_0}{k}(1 - e^{-kt/m})$$
Explanation
Position is the integral of velocity with respect to time. $$x(t) = \int_0^t v(\tau) d\tau = \int_0^t v_0 e^{-k\tau/m} d\tau$$. Evaluating the integral gives $$x(t) = v_0 [-\frac{m}{k}e^{-k\tau/m}]_0^t = -\frac{mv_0}{k}(e^{-kt/m} - e^0) = \frac{mv_0}{k}(1 - e^{-kt/m})$$.
A sphere of mass $$m$$ is released from rest in a viscous fluid. The fluid exerts a resistive force with magnitude $$F_r = kv$$, where $$v$$ is the speed of the sphere and $$k$$ is a positive constant. What is the terminal speed of the sphere?
$$\sqrt{mg/k}$$
$$mgk$$
$$k/mg$$
$$mg/k$$
Explanation
Terminal speed is reached when the net force on the sphere is zero. The forces acting on the sphere are gravity ($$mg$$ downward) and the resistive force ($$kv$$ upward). Setting the net force to zero: $$\sum F = mg - kv_T = 0$$. Solving for the terminal speed $$v_T$$ gives $$v_T = mg/k$$.
An object of mass $$m$$ is dropped from rest and experiences a drag force of magnitude $$F_r = kv$$, where $$v$$ is its speed and $$k$$ is a constant. Taking the downward direction as positive, which of the following differential equations correctly describes the object's motion?
$$m \frac{dv}{dt} = kv - mg$$
$$m \frac{dv}{dt} = mg + kv$$
$$m \frac{dv}{dt} = -kv$$
$$m \frac{dv}{dt} = mg - kv$$
Explanation
According to Newton's second law, $$F_{net} = ma$$. The net force is the sum of the gravitational force ($$mg$$, acting downward, which is the positive direction) and the drag force ($$kv$$, acting upward, which is the negative direction). Therefore, $$F_{net} = mg - kv$$. Since acceleration $$a = dv/dt$$, the equation of motion is $$m \frac{dv}{dt} = mg - kv$$.
Two small spheres, A and B, have the same radius and shape, but sphere A has mass $$M$$ and sphere B has mass $$2M$$. They are dropped from rest in air, and each experiences a drag force proportional to its speed, with the same proportionality constant $$k$$. Let $$v_{TA}$$ and $$v_{TB}$$ be their respective terminal speeds. Which of the following is true?
$$v_{TB} = 2v_{TA}$$
$$v_{TB} = v_{TA} / 2$$
$$v_{TB} = \sqrt{2}v_{TA}$$
$$v_{TB} = v_{TA}$$
Explanation
The terminal speed $$v_T$$ for an object with linear drag is given by $$v_T = mg/k$$. For sphere A, $$v_{TA} = Mg/k$$. For sphere B, the mass is $$2M$$, so its terminal speed is $$v_{TB} = (2M)g/k = 2(Mg/k) = 2v_{TA}$$. The more massive object has a higher terminal speed.
An object of mass $$m$$ is dropped from rest into a fluid with a resistive force $$F_r = kv$$. Its velocity as a function of time $$t$$ is given by which of the following expressions?
$$v(t) = \frac{mg}{k} e^{-kt/m}$$
$$v(t) = gt$$
$$v(t) = \frac{mg}{k} (1 - e^{-kt/m})$$
$$v(t) = \frac{mg}{k} (1 + e^{-kt/m})$$
Explanation
The differential equation is $$m \frac{dv}{dt} = mg - kv$$. Solving this equation with the initial condition $$v(0)=0$$ yields the solution $$v(t) = \frac{mg}{k}(1 - e^{-kt/m})$$. This function correctly shows the velocity starting at 0 and asymptotically approaching the terminal velocity $$v_T = mg/k$$.
An object is falling through a fluid and experiences a resistive force proportional to its velocity. The object reaches terminal velocity when which of the following conditions is met?
the magnitude of the resistive force equals the magnitude of the gravitational force.
the object has fallen for a time equal to the characteristic time constant $$m/k$$.
the object's acceleration is at its maximum possible value, $$g$$.
the object's kinetic energy is equal to the gravitational potential energy lost.
Explanation
Terminal velocity is the constant speed that a freely falling object eventually reaches when the resistance of the medium through which it is falling equals the force of gravity. At this point, the net force on the object is zero, and its acceleration becomes zero. Therefore, the magnitude of the resistive force must be equal to the magnitude of the gravitational force, $$mg$$.
An object is thrown downward with an initial speed greater than its terminal speed in a fluid with a resistive force proportional to speed. Which statement correctly describes the subsequent motion?
The object's speed increases until it reaches a new, higher terminal speed.
The object's speed decreases until it reaches its terminal speed.
The object's speed remains constant at its initial value because it is already moving fast.
The object immediately reverses direction and moves upward toward the surface.
Explanation
If the initial downward speed $$v_0$$ is greater than the terminal speed $$v_T$$, the initial upward resistive force $$kv_0$$ is greater than the downward gravitational force $$mg$$. This results in a net upward force, causing an upward acceleration (i.e., a deceleration). The object slows down, the resistive force decreases, and the speed approaches the terminal speed from above.
A motorboat of mass $$m$$ is propelled by a constant force $$F_{app}$$. The water exerts a resistive force of magnitude $$kv$$, where $$v$$ is the speed of the boat. Which differential equation describes the motion of the boat?
$$m \frac{dv}{dt} = kv - F_{app}$$
$$m \frac{dv}{dt} = F_{app} + kv$$
$$m \frac{dv}{dt} = F_{app} - kv$$
$$m \frac{dv}{dt} = -F_{app} - kv$$
Explanation
Using Newton's second law, $$F_{net} = ma$$. The net force is the sum of the applied force $$F_{app}$$ (in the direction of motion) and the resistive force $$kv$$ (opposite to the direction of motion). Thus, $$F_{net} = F_{app} - kv$$. Setting this equal to $$ma = m \frac{dv}{dt}$$ gives the correct differential equation.