down because of the traffic lights, so because F (^) net = ma and acceleration is large, the engine must
produce a lot of force. However, on the freeway, the car moves with constant velocity, and
acceleration is zero. So the engine produces no force, allowing for better gas mileage.
Ernie says: Gas mileage is better in town. In town, the speed of the car is slower than the speed on
the freeway. Acceleration is velocity divided by time, so the acceleration in town is smaller. Because
F (^) net = ma , then, the force of the engine is smaller in town giving better gas mileage.
Oscar says: Gas mileage is better on the freeway. The force of the engine only has to be enough to
equal the force of air resistance—the engine doesn’t have to accelerate the car because the car
maintains a constant speed. Whereas in town, the force of the engine must often be greater than the
force of friction and air resistance in order to let the car speed up.
Solutions to Practice Problems
1 . B— “Smooth” usually means, “ignore friction.” So the only force acting along the plane is a
component of gravity, mg (sin 30°). The F (^) net equation becomes mg (sin 30°) − 0 = ma . The mass
cancels, leaving the acceleration as 5 m/s^2 . What’s left is a kinematics problem. Set up a chart, calling
the direction down the plane as positive:
Use ** (Δx = v 0 t + ^1 / 2 at 2 ) to find that the time is 3.2 s.
2 . B— The normal force exerted on an object on an inclined plane equals mg (cos θ ), where θ is the
angle of the incline. If θ is greater than 0, then cos θ is less than 1, so the normal force is less than the
object’s weight.
3 . E— Consider the forces acting on each block separately. On the 1.0-kg block, only the tension acts, so
T = (1.0 kg)a . On the 2.0-kg block, the tension acts left, but the 10 N force acts right, so 10 N − T =
(2.0 kg)a . Add these equations together (noting that the tension in the rope is the same in both
equations), getting 10 N = (3.0 kg )a ; acceleration is 3.3 m/s^2 . To finish, T = (1.0 kg)a , so tension is
3.3 N.
4 . B— The acceleration is given by the slope of the v –t graph, which has magnitude 0.4 m/s^2 . F (^) net =
ma , so 5 kg × 0.4 m/s^2 = 2.0 N. This force is to the left because acceleration is negative (the slope is
negative), and negative was defined as left.
5 . The setup is the same as in the chapter’s example problem, except this time there is a force of friction
acting to the left on the 5-kg block. Because this block is in equilibrium vertically, the normal force is