An object is being lowered on a cord at a constant speed.

How does the tension T in the cord compare to the weight mg of the object?

Blue: T = mg        Yellow: T > mg        Green: T < mg
 

Answer: T=mg. Since velocity=const, that means a=0, and F_net=0, so the upward force T must just cancel the downward force mg.

An object is being lowered on a cord at a speed which is decreasing. There are two forces on the object, the weight, magnitude mg, and the tension, magnitude T, in the cord.

Which equation is true?

Blue: T = mg        Yellow: T > mg        Green: T < mg
 

Answer. T>mg. Since the object is accelerating, there must be a net force, so we cannot have T=mg. (more below).

HINT: What is the direction of the acceleration?

Pink: up         Yellow: down         Green: a=0
 

Answer: acceleration is up, since up is the direction of .
 
 

It is usually a good idea to choose the direction of the acceleration as the (+) direction. Choosing up as the (+) direction, we can write Since a>0, we have (T-mg)>0 and T>mg.

If we had chosen down as the (+) direction, then we would write  But a and F_net would turn out to be negative (since both are in the upward direction) and we would have F_net=(mg-T) < 0, and it would still be true that T>mg.

Situation I: A constant force is exerted for a short time interval on a frictionless cart that is initially at rest. The cart acquires a final velocity v_f. (1D Motion)

Situation II: The same constant force is exerted for the same short time interval on a frictionless cart that is initially moving at velocity v₁. The cart has a final velocity of v₂ . The change in the speed of the cart D v = v₂ - v₁ , compared to the final speed in situation I, is..

Blue: the same, D v = v_f.         Pink: greater, D v > v_f.

Green: less, D v < v_f.        Yellow: answer depends on the sign of v₁ and v₂.

Answer: D v = v_f. For both situation I and II, we have . (Here the motion is 1D, so we can drop the arrow signs over the vectors.) For both situations I and I, the three quantities F_net, m, and D t are identical. So Dv must be the same also. For situation I, D v=(v_f-0)=v_f.

A car of mass m, traveling at constant speed, rides over the top of a hill.

The magnitude of the normal force N of the road on the car is..

Blue: greater than the weight of the car, N > mg.

Yellow: equal to the weight, N = mg.        Green: less than the weight, N < mg.
 

Answer: N < mG !   The car is moving along a circular path, so it must be accelerating toward the center of the circle (which is down when the car is at the top of the hill ). If there is an acceleration downward, there must be a net force downward (since ). Since there is a net force downward, the downward force (mg) must have greater magnitude than the upward force (N). If the car were parked at the crest of the hill, then we would have N=mg. If the car goes over the hill several times, going faster and faster each time, the normal force gets smaller and smaller, until the normal force goes to zero (at which point the car becomes airborne).