MOTION WITH A VARIYING FORCE
An air-track glider of mass 0.100 kg is attached to the end of a horizontal air track by a spring with force constant 20.0 N/m (Fig. 6.22a). Initially the spring is unstretched and the glider is moving at 1.50 m/s to the right. Find the maximum distance d that that the glider moves to the right (a) if the air track is turned on, so that there is no friction, and (b) if the air is turned off, so that there is kinetic friction with coefficient \mu_{\mathrm{k}} = 0.47.
IDENTIFY and SET UP:
The force exerted by the spring is not constant, so we cannot use the constant-acceleration formulas of Chapter 2 to solve this problem. Instead, we’ll use the work– energy theorem, since the total work done involves the distance moved (our target variable). In Figs. 6.22b and 6.22c we choose the positive x-direction to be to the right (in the direction of the glider’s motion). We take x = 0 at the glider’s initial position (where the spring is unstretched) and x = d (the target variable) at the position where the glider stops. The motion is purely horizontal, so only the horizontal forces do work. Note that Eq. (6.10) (W=\int_{x_{1}}^{x_{2}} F_{x} d x=\int_{x_{1}}^{x_{2}} k x d x=\frac{1}{2} k x_{2}^{2}-\frac{1}{2} k x_{1}^{2}) gives the work done by the glider on the spring as it stretches; to use the work–energy theorem we need the work done by the spring on the glider, which is the negative of Eq. (6.10). We expect the glider to move farther without friction than with friction.
EXECUTE:
(a) Equation (6.10) (W=\int_{x_{1}}^{x_{2}} F_{x} d x=\int_{x_{1}}^{x_{2}} k x d x=\frac{1}{2} k x_{2}^{2}-\frac{1}{2} k x_{1}^{2}) says that as the glider moves from x_1 = 0 to x_2 = d, it does an amount of work W=\frac{1}{2} k d^{2}-\frac{1}{2} k(0)^{2} = \frac{1}{2} k d^{2} on the spring. The amount of work that the spring does on the glider is the negative of this, –\frac{1}{2} k d^{2}, The spring stretches until the glider comes instantaneously to rest, so the final kinetic energy K_2is zero. The initial kinetic energy is \frac{1}{2} m v_{1}^{2} where v_1 = 1.50 m/s is the glider’s initial speed. From the work–energy theorem,
-\frac{1}{2} k d^{2}=0-\frac{1}{2} m v_{1}^{2}We solve for the distance d the glider moves:
d=v_{1} \sqrt{\frac{m}{k}}=(1.50 \mathrm{~m} / \mathrm{s}) \sqrt{\frac{0.100 \mathrm{~kg}}{20.0 \mathrm{~N} / \mathrm{m}}} = 0.106 m = 10.6 cm
The stretched spring subsequently pulls the glider back to the left, so the glider is at rest only instantaneously.
(b) If the air is turned off, we must include the work done by the kinetic friction force. The normal force n is equal in magnitude to the weight of the glider, since the track is horizontal and there are no other vertical forces. Hence the kinetic friction force has constant magnitude f_{\mathrm{k}}=\mu_{\mathrm{k}} n=\mu_{\mathrm{k}} m g. The friction force is directed opposite to the displacement, so the work done by friction is
W_{\text {fric }}=f_{\mathrm{k}} d \cos 180^{\circ}=-f_{\mathrm{k}} d=-\mu_{\mathrm{k}} m g dThe total work is the sum of Wfric and the work done by the spring,
– \frac{1}{2} k d^{2}. The work–energy theorem then says that
-\mu_{\mathrm{k}} m g d-\frac{1}{2} k d^{2}=0-\frac{1}{2} m v_{1}^{2} or
\frac{1}{2} k d^{2}+\mu_{\mathrm{k}} m g d-\frac{1}{2} m v_{1}^{2}=0This is a quadratic equation for d. The solutions are
d=-\frac{\mu_{\mathrm{k}} m g}{k} \pm \sqrt{\left(\frac{\mu_{\mathrm{k}} m g}{k}\right)^{2}+\frac{m v_{1}^{2}}{k}}We have
\frac{\mu_{\mathrm{k}} m g}{k}=\frac{(0.47)(0.100 \mathrm{~kg})\left(9.80 \mathrm{~m} / \mathrm{s}^{2}\right)}{20.0 \mathrm{~N} / \mathrm{m}}=0.02303 \mathrm{~m}\frac{m v_{1}^{2}}{k}=\frac{(0.100 \mathrm{~kg})(1.50 \mathrm{~m} / \mathrm{s})^{2}}{20.0 \mathrm{~N} / \mathrm{m}}=0.01125 \mathrm{~m}^{2}
so d=-(0.02303 \mathrm{~m}) \pm \sqrt{(0.02303 \mathrm{~m})^{2}+0.01125 \mathrm{~m}^{2}} = 0.086 m or -0.132 m
The quantity d is a positive displacement, so only the positive value of d makes sense. Thus with friction the glider moves a distance d = 0.086 m = 8.6 cm.
EVALUATE: we set \mu_{\mathrm{k}} = 0, our algebraic solution for d in part (b) reduces to d=v_{1} \sqrt{m / k}, the zero-friction result from part (a).
With friction, the glider goes a shorter distance. Again the glider stops instantaneously, and again the spring force pulls it toward the left; whether it moves or not depends on how great the static friction force is. How large would the coefficient of static friction \mu_{\mathrm{s}} have to be to keep the glider from springing back to the left?