Section Test:Force and Motion

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Topics Covered

This Section Test deals with the basic force topics. Force diagrams, forces due to gravity, springs, friction, and centripetal force are all covered.


Section Test

1. A bowling ball of mass 2 kg is dropped from an airplane and begins to accelerate at 9.8m/s2, after a short time the acceleration of the bowling ball is zero, which of the following is true in regards to the bowling ball?

The bowling ball no longer experiences a downward pulling force.
The bowling ball's net force is 19.6 kgm/s2 downward.
The bowling ball is experiencing a force of 19.6 kgm/s2 upward.
The bowling ball is experiencing no forces because it has no acceleration.
→ To answer this problem we must understand what forces are acting on the ball. Initially, when the ball is first dropped gravity is the only force acting, and it is pulling the ball down, thus at this initial point, it's net force would also be equal to it, because it is the only force! After some time however, the net acceleration becomes zero, which implies that the net force is zero, this does not however imply that there are no forces acting, only that in some way they sum to zero. Because gravity is pulling down always, there must be a second force (in our case it will be air friction) which begins to push upwards (in the opposite direction of gravity) and because the acceleration is now zero, it must be equal to gravity's force down. The force of gravity down is F = ma = (2kg)(9.8m/s2), which must also equal the force upwards.

2. A mass exerting a force north at 10N experiences a second force of 10N towards the west. What is the net force of the block?

10N northwest
14N northwest
20N northwest
34N northwest
→ Adding forces isn't as simple as just adding their values together, they are vectors, and thus their vector components must be taken into account. 10N north and 10N west does not add to 10 in some intermediate direction, we must calculate it. For two vectors, the sum of them will be equal to \sqrt[2]{(\text{magnitude of vector one})^{2} + (\text{magnitude of vector two})^{2}}. In our case this ends up equaling about 14. The direction must also generally be solved for, but thankfully in this problem we know one is north, the other is west, both are the same size, thus it will be the average direction between the two, or northwest.

3. A 0.10 kg hockey puck is set in motion across a frozen pond. If ice friction is neglected, the force from a hockey stick required to keep the puck sliding at a constant velocity would be?

0.0 N
0.98 N
9.8 N
98.0 N
→ Any object will remain in motion unless acted upon by an external force, this is one of Newton's laws! Thus if a puck is sliding without friction it will keep sliding forever without the need of any force.

4. A proton being pushed by a magnetic field experiences a force of 100N. A second force of 100N is suddenly applied in the reverse direction. The proton will?

be brought to a sudden halt
decelate gradually to a halt
continue at the speed it had when it encountered the second force
none of these
→ When the first force is applied alone, the net force is equal to this and the proton accelerates, when the second force applied it cancels the first force out (because they are equal and opposite) and thus the proton no longer has a net force, and thus no longer has acceleration. As a consequence, the proton will simply continue traveling at whatever velocity it had, without acceleration or deceleration.

5. Twin-jet engines accelerate a 900kg fighter plane to 240 m/s from rest in 60 seconds, what force must each engine be generating?

1800 N
3600 N
4200 N
5600 N
→ To determine the force, we must first determine the acceleration. Using kinematics equations we can easily find this. v_{f} = v_{i} + at or, a = \frac{v_{f} - v_{i}}{t} = \frac{240 - 0}{60} = 4. Now that we have the acceleration, we can find the total force, ma = 900kg * 4m/s2 = 3600N. There are however, two engines, thus the total force is split between the two of them and each yields 1800N.

6. A cart on a roller coaster follows a vertical circular loop of diameter 18m in which at the top of the loop passengers experience a feeling of weightlessness. What must the velocity of the roller coaster be at the top of the track for this to occur?

3 m/s
5 m/s
9 m/s
12 m/s
→ We can draw a force diagram for the cart at the top of the loop. For a coaster on the loop, there will be gravity pulling the cart down, a normal force pushing the cart down (because the tracks are above it and it pushs out from the tracks). Thus the net force = mg + N = ma. For the cart to experience weightlessness, the cart can't be pushing into the tracks at all (or falling off) thus the normal force N, will be zero. In this case, mg = ma, so the cart's acceleration inwards must equal its gravitational force. To determine the velocity needed we need only to recognize that Fnet is the centripital force inwards, and thus the acceleration inwards = v2/r. Therefore ma = mv2/r, or v2 = ar = (9.8)(9) ~= 81. Therefore the velocity must be 9m/s.

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