Physics 2.1 Forces and their effects

What is acceleration?

What is the resultant force?

If a non-zero resultant force acts on a stationary object?

If a resultant force acting on a stationary object is zero it will remain stationary?

If the resultant force acting on a moving object is zero?

If the resultant force acting on an object is non-zero, it will accelerate in the direction of the resultant force?

When the resultant force acts on an object, the object can experience 5 forms of acceleration: [blank_start]speeding up,[blank_end] [blank_start]slowing down[blank_end], [blank_start]starting[blank_end], [blank_start]stopping[blank_end] and [blank_start]changing direction[blank_end]

How do you calculate the acceleration produced by a resultant force?

When two objects interact, they exert forces on each other that are [blank_start]equal[blank_end] and [blank_start]opposite[blank_end]

On the surface of the planet, gravity makes everything [blank_start]accelerate[blank_end] towards the [blank_start]ground[blank_end] and gives everything a [blank_start]weight[blank_end]

To calculate the weight of an object using its mass and gravitational field strength: [blank_start]W=mxg.[blank_end] W is [blank_start]weight in N[blank_end] M is [blank_start]mass in Kg[blank_end] G is [blank_start]gravitational field strength in N/KG[blank_end]

For a vehicle to travel at a steady speed...

The [blank_start]frictional[blank_end] forces that oppose the motion of an object moving through a fluid [blank_start]increase[blank_end] with the objects [blank_start]speed[blank_end].

Ways of increasing a vehicle's top speed include: * [blank_start]Reducing drag[blank_end]: by altering the [blank_start]shape[blank_end] of a vehicle to make it more [blank_start]streamlined[blank_end]. * Increasing the [blank_start]power of the vehicle's engine[blank_end]: meaning the [blank_start]driving force[blank_end] becomes [blank_start]larger[blank_end] so the drag force on the vehicle will [blank_start]equal the driving force[blank_end] at a [blank_start]higher[blank_end] speed.

What is terminal velocity?

When an object falls through a fluid it initially [blank_start]accelerates[blank_end] due to the force of [blank_start]gravity[blank_end] being [blank_start]greater[blank_end] than the [blank_start]frictional[blank_end] forces. As the object moves [blank_start]faster[blank_end], the [blank_start]frictional forces[blank_end] that act on it [blank_start]increase[blank_end]. This gradually [blank_start]reduces[blank_end] the [blank_start]acceleration[blank_end] until the [blank_start]frictional[blank_end] force is [blank_start]equal[blank_end] to the [blank_start]accelerating[blank_end] force so it wont [blank_start]accelerate[blank_end] anymore. It will have reached its [blank_start]maximum speed[blank_end] ([blank_start]terminal velocity[blank_end]) and will fall at a [blank_start]steady rate[blank_end].

The terminal velocity of falling objects depends on their [blank_start]shape[blank_end] and [blank_start]area[blank_end]. The accelerating force acting on all objects is [blank_start]gravity[blank_end] and if it were not for [blank_start]air resistance[blank_end], everything would fall at the [blank_start]same[blank_end] rate (like on the [blank_start]moon[blank_end]). The terminal velocity of objects is determined by its [blank_start]drag[blank_end] in comparison to its [blank_start]weight[blank_end]. The frictional force depends on its shape and area.

A parachute helps to reduce a diver's terminal velocity because it [blank_start]increases[blank_end] the surface area which increases the [blank_start]air resistance[blank_end] acting on the diver. This causes the diver to [blank_start]slow down[blank_end], which [blank_start]reduces[blank_end] his speed and therefore the [blank_start]terminal velocity[blank_end].

What is the stopping distance?

Thinking distance is the distance the vehicle travels during the driver's [blank_start]reaction[blank_end] time Braking distance is the [blank_start]distance[blank_end] the vehicle travels after the [blank_start]brakes[blank_end] are applied until it comes to a complete [blank_start]stop[blank_end]

For a given braking force... what happens to the stopping distance?

Factors that affect thinking distance: * How [blank_start]fast[blank_end] you're going - the faster you go, the [blank_start]longer[blank_end] the stopping distance * How [blank_start]quick[blank_end] to [blank_start]respond[blank_end] you are - this can be affected by [blank_start]tiredness[blank_end], [blank_start]drugs[blank_end] and [blank_start]alcohol[blank_end] * Bad [blank_start]visibility[blank_end] *[blank_start]Distractions[blank_end] (e.g. rain, noise)

What does NOT affect braking distance.

Braking and kinetic energy transfer: To slow down a car, [blank_start]kinetic[blank_end] energy needs to be [blank_start]converted[blank_end] into other types of [blank_start]energy[blank_end]. (e.g.thermal and a little sound) When the brakes of a car are applied, the [blank_start]friction[blank_end] between the [blank_start]wheels[blank_end] and the [blank_start]brakes[blank_end] converts kinetic energy into thermal energy, which causes the temperature of the brakes to [blank_start]increase[blank_end]. So, work is done by the [blank_start]braking[blank_end] force to convert the kinetic energy into thermal energy and a little [blank_start]sound[blank_end] energy.

What will NOT happen to an object when a force is applied to it?

Is an elastic object one that can change shape under an applied force, and then return to its original shape after the force has been removed?

When a force does work to change the shape of an elastic object, the object [blank_start]stretches[blank_end] and stores [blank_start]elastic potential energy[blank_end]. The [blank_start]elastic potential energy[blank_end] is then [blank_start]converted[blank_end] to [blank_start]kinetic[blank_end] energy when the force is [blank_start]removed[blank_end] and the object returns to its [blank_start]original[blank_end] shape.

When a spring is supported at the top then a weight is attached to the bottom, it stretches. What is the formula for calculating the force applied to an object, given its extension and spring constant?

Hooke's law: The extension of an elastic object is [blank_start]directly proportional[blank_end] to the [blank_start]force applied[blank_end] until the [blank_start]limit of proportionality[blank_end] is suceeded

Is the limit of proportionality the point in which the an elastic object will no longer extend proportionally with the applied force?