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PHYSICS, M2 EQ-Bank 3

A horizontal Atwood machine begins accelerating from rest along a horizontal track of length, \(L=4.00\ \text{m}\).

A block with a mass of 500 g takes 2.7 seconds to travel the full length of the track. A constant frictional force acts on the block, described by the equation \(F_f = \mu_k mg\)
  

  1. Determine the coefficient of kinetic friction between the 500 g block and the track.   (4 marks)
  1. Find the kinetic energy of the 500 g block when it reaches the end of the track.   (2 marks)
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a.    \(0.154\)

b.    \(2.194\ \text{J}\)

Show Worked Solution

a.     Find system acceleration using  \(s=ut + \dfrac{1}{2}at^2\):

\(s\) \(=ut +\dfrac{1}{2}at^2\)  
\(4.00\) \(=0 + \dfrac{1}{2} \times a \times 2.7^2\)  
\(a\) \(=\dfrac{2 \times 4.00}{2.7^2}=1.097\ \text{ms}^{-2}\)  

 
\(\therefore\) Total system mass \((m_T)\) = 500 + 150=650\ \text{g} = 0.65\ \text{kg} \)
 

Find frictional force acting on the block:

\(F_{\text{net}}\) \(=F_{\text{applied}}-F_f\)  
\(0.65 \times 1.097\) \(= 0.15 \times 9.8-F_f\)  
\(F_f\) \(=1.47-0.713=0.757\ \text{N}\)  

 
\(\therefore \mu_k = \dfrac{F_f}{mg} = \dfrac{0.757}{9.8 \times 0.5} = 0.154\)
  

b.    Using  \(v^2=u^2 +2as:\)

\(v^2=u^2+2as = 0^2+2 \times 1.097 \times 4.00 = 8.776\ \text{ms}^{-1}\)

\(KE = \dfrac{1}{2}mv^2 = \dfrac{1}{2} \times 0.5 \times 8.776 = 2.194\ \text{J}\)

PHYSICS, M2 EQ-Bank 7 MC

A large crate with a mass of 180 kg is being dragged across a warehouse floor using a horizontal rope. The tension in the rope is 500 N, and the crate accelerates at a rate of 2.0 ms\(^{-2}\).
 

The magnitude of the frictional force acting between the floor and the crate is:

  1. \(140\ \text{N}\)
  2. \(180\ \text{N}\)
  3. \(360\ \text{N}\)
  4. \(0\ \text{N}\)
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\(A\)

Show Worked Solution
  • \(\vec{F}_{\text{net}} = ma = 180 \times 2 = 360\ \text{N}\)
  • \(F_f = T-F_{\text{net}} = 500-360=140\ \text{N}\)

\(\Rightarrow A\)

PHYSICS, M2 EQ-Bank 2

Consider the information in the diagram below.
  
  1. Calculate the acceleration of the blocks as a result of the applied forces acting on them.   (2 marks)
  1. What forces does block \(Y\) apply to block \(X\).   (2 marks)
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a.    \(0.5\ \text{ms}^{-2}\)

b.    \(1500\ \text{N}\)

Show Worked Solution

a.    Total system mass \(= 1000 + 2000 = 3000\ \text{kg}\)

Applied force on system \(= 2250\ \text{N}\)

Frictional force on system \(= 0.25 \times 3000 = 750\ \text{N}\)

Net force \(=\) the applied force \(-\) the frictional force \(=2250-750 =1500\ \text{N}\)

Find acceleration using \(F_{\text{net}}= ma\):

\(a=\dfrac{F_{\text{net}}}{m} =\dfrac{1500}{3000} = 0.5\ \text{ms}^{-2}\)

 

b.    Net force on \(X = ma = 1000 \times 0.5 = 500\ \text{N}\) 

Net force on \(X\)  \(=\)  applied force on \(X -\) force of \(Y\) on \(X -\) the frictional force on \(X\).  
\(500\) \(=2250-F_{Y → X}-(1000 \times 0.25)\)  
\(F_{Y → X}\) \(=2250-500-250=1500\ \text{N}\)  

PHYSICS, M2 EQ-Bank 1

A delivery person is pushing a 25 kg cart along a flat horizontal floor. They apply a force of 50 N on the handle, directed at an angle of 30° above the horizontal as shown below.
 

  1. What is the magnitude of the normal force exerted by the floor on the cart?   (2 marks)
  1. If the cart moves at a constant speed, what is the frictional force opposing its motion?   (1 mark)
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a.    \(270\ \text{N}\)

b.    \(43.3\ \text{N}\)

Show Worked Solution

a.    Normal force:

\(F_N\) \(=mg + F\sin \theta\)  
  \(=25 \times 9.8 + 50 \times \sin 30^{\circ}\)  
  \(=270\ \text{N}\)  

 

b.    Cart moves at a constant speed:

  • Applied horizontal force = frictional force opposing the motion.
  •    \(F_f= F \cos \theta = 50 \times \cos 30^{\circ} = 43.3\ \text{N}\).

PHYSICS, M2 EQ-Bank 6 MC

A box of mass \(m\) is sliding across a horizontal surface and slowing down with a constant deceleration of 3.0 ms\(^{-2}\).
 

What is the coefficient of kinetic friction, \(\mu\), closest to?

  1. \(29.4\)
  2. \(3.1\)
  3. \(0.31\)
  4. \(0.15\)
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\(C\)

Show Worked Solution
  • The coefficient of kinetic friction can be calculated using the equation \(F_f=\mu F_N\):
\(F_f\) \(=\mu F_N\)  
\(ma\) \(=\mu \times mg\)  
\(a\) \(=\mu \times g\)  
\(\mu\) \(=\dfrac{a}{g}=\dfrac{3}{9.8}=0.31\)  

 
\(\Rightarrow C\)

PHYSICS, M2 EQ-Bank 1 MC

A block with a mass of 300 kg is sliding across a frictionless floor at a constant speed of 6 ms\(^{-1}\).

What magnitude of force is required to keep the block moving at this constant speed?

  1. \(0\ \text{N}\)
  2. \(0.03\ \text{N}\)
  3. \(50\ \text{N}\)
  4. \(1800\ \text{N}\)
Show Answers Only

\(A\)

Show Worked Solution
  • The object is moving at a constant velocity. 
  • By Newton’s first law of motion, an object travelling at a constant velocity will continue to travel at a constant velocity unless acted upon by an unbalanced force. 
  • As the block is moving on a frictionless surface, there is no net force on the block. The block requires \(0\ \text{N}\) of force to continue to move at a constant speed.

\(\Rightarrow A\)

PHYSICS, M2 2023 VCE 8

Maia is at a skatepark. She stands on her skateboard as it rolls in a straight line down a gentle slope at a constant speed of 3.0 m s\(^{-1}\), as shown in the figure below. The slope is 5° to the horizontal.

The combined mass of Maia and the skateboard is 65 kg.
  

  1. The combined system of Maia and the skateboard is modelled as a small box with point \(\text{M}\) at the centre of mass, as shown below.
  2. Draw and label arrows to represent each of the forces acting on the system - that is, Maia and skateboard as they roll down the slope.   (3 marks)
     


  1. Calculate the magnitude of the total frictional forces acting on Maia and the skateboard.  (2 marks)

Near the bottom of the ramp, Maia takes hold of a large pole and comes to a complete rest while still standing on the skateboard. Maia and the skateboard now have no momentum or kinetic energy.

  1. Explain what happened to both the momentum and the kinetic energy of Maia and the skateboard. No calculations are required.  (2 marks)

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a.
         

b.     55.5 N

c.    Momentum:

  • Maia and the skateboards momentum was transferred into the pole and hence into the Earth. Due to the Earth’s very large mass, the effect on its velocity is negligible.

Kinetic energy:

  • The kinetic energy of Maia and the skateboard would have been transformed into heat energy between the contact of Maia and the pole and/or transferred into the work done by Maia’s muscles in slowing herself down.

Show Worked Solution

a.
        

b.     Total frictional forces:

  • Constant speed means that the force down the slope of the incline is equal to the sum of the frictional forces acting on Maia and the skateboard.
\(F_{\text{down slope}}\) \(=F_f\) \(=mg\, \sin\, \theta\)
    \(=65 \times 9.8 \times \sin\,5^{\circ}\)
    \(=55.5\ \text{N}\)
♦ Mean mark (b) 50%.

c.    Momentum:

  • Maia and the skateboards momentum was transferred into the pole and hence into the Earth. Due to the Earth’s very large mass, the effect on its velocity is negligible.

Kinetic energy:

  • The kinetic energy of Maia and the skateboard would have been transformed into heat energy between the contact of Maia and the pole and/or transferred into the work done by Maia’s muscles in slowing herself down.
♦♦♦ Mean mark (c) 26%.

PHYSICS, M2 2019 VCE 11 MC

An ultralight aeroplane of mass 500 kg flies in a horizontal straight line at a constant speed of 100 ms\(^{-1}\).

The horizontal resistance force acting on the aeroplane is 1500 N.

Which one of the following best describes the magnitude of the forward horizontal thrust on the aeroplane?

  1. 1500 N
  2. slightly less than 1500 N
  3. slightly more than 1500 N
  4. 5000 N
Show Answers Only

\(A\)

Show Worked Solution
  • There is no net force acting on the aeroplane as it is travelling at a constant speed.
  • The forward horizontal thrust on the plane must be opposite in direction but equal in magnitude to the horizontal resistance force acting on the plane.

\(\Rightarrow A\)

PHYSICS, M2 2022 VCE 7

Kym and Kelly are experimenting with trolleys on a ramp inclined at 25°, as shown in the diagram below. They release a trolley with a mass of 2.0 kg from the top of the ramp. The trolley moves down the ramp, through two light gates and onto a horizontal, frictionless surface. Kym and Kelly calculate the acceleration of the trolley to be \(3.2\ \text{m s}^{-2}\) using the information from the light gates.
 
 


 

  1.  i. Show that the component of the gravitational force of the trolley down the slope is \(8.3 \text{ N}\). Use \(g=9.8 \text{ m s}^{-2}\).  (2 marks)
  1. ii. Assume that on the ramp there is a constant frictional force acting on the trolley and opposing its motion.
  2.      Calculate the magnitude of the constant frictional force acting on the trolley.  (2 marks)
  1. When it reaches the bottom of the ramp, the trolley travels along the horizontal, frictionless surface at a speed of \(4.0\ \text{m s}^{-1}\) until it collides with a stationary identical trolley. The two trolleys stick together and continue in the same direction as the first trolley.
    1. Calculate the speed of the two trolleys after the collision. Show your working and clearly state the physics principle that you have used.  (3 marks)
    1. Determine, with calculations, whether this collision is an elastic or inelastic collision. Show your working.  (3 marks)
Show Answers Only

a.i.   See Worked Solutions

a.ii.  \(F_f=1.9\ \text{N}\)

b.i.   \(2.0\ \text{ms}^{-1}\)

 b.ii.  For the collision to be elastic, the kinetic energy must be conserved.

\(KE_{\text{init}}=\dfrac{1}{2} mv^2=\dfrac{1}{2} \times 2 \times 4^2=16\ \text{J}\)

\(KE_{\text{final}}=\dfrac{1}{2} mv^2=\dfrac{1}{2} \times 4 \times 2^2=8\ \text{J}\)

\(\therefore\) Since the kinetic energy of the system decreases after the collision, it is not an elastic collision.

Show Worked Solution

a.i.  The gravitational force down the slope:

\(F=mg\, \sin \theta=2.0 \times 9.8 \times \sin 25=8.3\ \text{N}\) 

 
a.ii. \(F_{net}=ma=2.0 \times 3.2=6.4\ \text{N}\)

\(6.4\) \(=F-F_f\)  
\(F_f\) \(=8.3-6.4\)  
  \(=1.9\ \text{N}\)  
♦ Mean mark (a.ii) 53%.

b.i.   By the conservation of momentum:

\(m_1u_1+m_2u_2\) \(=v(m_1+m_2)\)  
\(2 \times 4 + 2 \times 0\) \(=v(2 +2)\)  
\(8\) \(=4v\)  
\(v\) \(=2\ \text{ms}^{-1}\)  
 

 b.ii.  For the collision to be elastic, the kinetic energy must be conserved.

\(KE_{\text{init}}=\dfrac{1}{2} mv^2=\dfrac{1}{2} \times 2 \times 4^2=16\ \text{J}\)

\(KE_{\text{final}}=\dfrac{1}{2} mv^2=\dfrac{1}{2} \times 4 \times 2^2=8\ \text{J}\)

\(\therefore\) Since the kinetic energy of the system decreases after the collision, it is not an elastic collision.