IB Physics Topic 3.2 Notes Quiz

Look at the diagram of a block on a table. What is the net force acting on the block if it is at rest?

• The net force is zero because the upward normal force balances the downward gravitational force.

• The net force is equal to the weight of the block acting downward.

• The net force is equal to the normal force acting upward.

• The net force is greater than zero because the block is on a surface.

Look at the diagram of an object in free fall. What is the only force acting on the object?

• The only force acting on the object is gravity (gravitational force).

• The only force acting on the object is air resistance.

• The only force acting on the object is friction.

• The only force acting on the object is magnetic force.

What is the weight and mass of a 60 kg person on Earth? What if they were on the moon?

• On Earth: Mass = 60 kg, Weight = 60 kg × 9.8 N/kg = 588 N. On the Moon: Mass = 60 kg, Weight = 60 kg × 1.6 N/kg = 96 N.

• On Earth: Mass = 60 kg, Weight = 60 kg × 10 N/kg = 600 N. On the Moon: Mass = 60 kg, Weight = 60 kg × 1.6 N/kg = 120 N.

• On Earth: Mass = 60 kg, Weight = 60 kg × 9.8 N/kg = 588 N. On the Moon: Mass = 60 kg, Weight = 60 kg × 1.6 N/kg = 80 N.

• On Earth: Mass = 60 kg, Weight = 60 kg × 9.8 N/kg = 588 N. On the Moon: Mass = 60 kg, Weight = 60 kg × 1.6 N/kg = 160 N.

A 10 kg block sits on a flat rough surface. If 25 N of force is required to make the box accelerate, what is the coefficient of friction between the block and the surface?

• 0.255

• 0.40

• 0.10

• 0.98

A 40 kg sled is sliding across a flat snow path at a constant speed. If the coefficient of friction between the snow and the sled is 0.05, what is the force applied to the sled?

• 9.8 N

• 19.6 N

• 2 N

• 40 N

A piece of wood has a volume of 34 cm³ and a mass of 29 g. Calculate the buoyancy force the wood will experience if it is completely submerged in water.

• 0.333 N

• 0.245 N

• 0.500 N

• 0.100 N

Assuming the force holding the wood underwater is removed so the wood is able to float in the water. Determine the volume of wood that will be floating above the surface. (Density of water = 1.0 g/cm³ = 1000 kg/m³, One Liter of water = 1000 cm³ = 1.00 kg)

• 5 cm³

• 2 cm³

• 10 cm³

• 0.5 cm³

A sphere of radius 3.3 cm travels through a fluid of viscosity 5.6 kg m⁻¹ s⁻¹. The viscous drag force is 15 N. Calculate the velocity of the sphere. (Use F_d = 6πηrv)

• 0.041 m/s

• 0.12 m/s

• 0.0041 m/s

• 0.82 m/s

A 90 kg skydiver jumps out of a plane and accelerates downward with an initial acceleration of 8.5 m/s². a. Determine the amount of friction acting on the skydiver when he jumps out of the plane. (Show your calculation.)

• Friction (air resistance) = Weight - Net force Weight = mass × gravity = 90 kg × 9.8 m/s² = 882 N Net force = mass × acceleration = 90 kg × 8.5 m/s² = 765 N Friction = Weight - Net force = 882 N - 765 N = 117 N

• Friction (air resistance) = Weight + Net force Weight = mass × gravity = 90 kg × 9.8 m/s² = 882 N Net force = mass × acceleration = 90 kg × 8.5 m/s² = 765 N Friction = Weight + Net force = 882 N + 765 N = 1647 N

• Friction (air resistance) = Net force - Weight Weight = mass × gravity = 90 kg × 9.8 m/s² = 882 N Net force = mass × acceleration = 90 kg × 8.5 m/s² = 765 N Friction = Net force - Weight = 765 N - 882 N = -117 N

• Friction (air resistance) = mass × acceleration only = 90 kg × 8.5 m/s² = 765 N

As the skydiver continues to fall toward the ground, his acceleration will:

• increase continuously as he falls.

• remain constant throughout the fall.

• decrease until it reaches zero at terminal speed.

• increase and then remain constant.