Question 2.5: Additional Answer Options

2021 - November - Paper 1 - Question 2

2.1

A body will remain in its state of rest or motion at constant velocity unless a non-zero resultant/net force/unbalanced force acts on it.

A body will remain in its state of rest or uniform motion in a straight line unless a (non-zero) resultant/net/unbalanced force acts on it.

2 marks

(2 x 1 mark)

2.2

w	F_g / F_w / weight/mg/196 N / gravitational force
F	F_A / Applied force
f_k	(kinetic) Friction /F_f/ f /18 N / F_w / f_w
N	F_N / Normal / 169,74 N

It is very important that you draw the free body diagram on an incline, therefore, the forces other than gravity must act parallel and perpendicularly to the incline.

4 marks

(4 x 1 mark)

Marking criteria:

Mark awarded for label and arrow, but penalise only once if arrows are omitted.
Do not penalise for the length of arrows, the drawing is not to scale.
For any other additional force(s) deduct 1 mark.
If force(s) do not make contact with the body deduct 1 mark.

2.3

How to approach this question:

The object is moving at a constant velocity, therefore, it must have an acceleration of 0 m·s^-2.

Option 1:

$\left.\begin{array}{l}F_{net}=ma\\ F+f_{k}+w_{\|l}=ma\\ F+\left(-f_{k}\right)+\left(-w_{\|}\right)=ma\\ F-\left(f_{k}+w_{\|}\right)=ma\end{array}\right\}Anyone$

$F-\left[18+(20)(9,8)\left(\sin30^{\circ}\right)\right]=0$

$F=116\mathrm{~N}$

Remember, the frictional force always acts in the opposite direction to the direction of motion.

Option 2:

Using the work-energy theorem, since the block travels at a constant velocity the change in kinetic energy is zero.

$W_{net}=\Delta E_{k}$

$F\Delta x\cos0^{\circ}+f\Delta x\cos180^{\circ}+w\Delta x\cos120^{\circ}=0$

$F\Delta x=18\Delta x+(20)(9,8)\Delta x(0,5)$

$F=116\mathrm{~N}$

4 marks

(4 x 1 mark)

2.4

116 N OR f + w||

Down the incline/opposite to the direction of motion.

The resultant force when the force F is removed, it will be equal to the sum of the weight and the frictional force acting down the incline.

2 marks

(2 x 1 mark)

2.5

How to approach this question:

If down the slope is taken as the positive direction.

Option 1:

Part 1:

$F_{net}=ma$

116=20a

$a=5,80\mathrm{~m}\cdot\mathrm{s}^{-2}$

Part 2:

The initial velocity is 2 m·s^-1up the slope.

$v_{f}=v_{i}+a\Delta t$

$0=-2+(5,8)\Delta t$

$\Delta t=0,34\mathrm{~s}$

$F_{net}\Delta t=m\left(v_{f}-v_{i}\right)$

$(116)\Delta t=(20)(0-(-2))$

$\Delta\mathrm{t}=0,34\mathrm{~s}$

$v_{f}=v_{i}+a\Delta t$

$0=-2+(5,8)\Delta t$

$\Delta t=0,34\mathrm{~s}$

$\mathrm{F}_{\mathrm{net}}\Delta\mathrm{t}=\mathrm{m}\left(\mathrm{v}_{\mathrm{f}}-\mathrm{v}_{\mathrm{i}}\right)$

$(116)\Delta t=(20)(0-(-2))$

$\Delta t=0,34\mathrm{~s}$

Part 3:

Since the direction of down the slope is positive the displacement will be negative.

$\Delta\mathrm{x}=\left(\frac{\mathrm{v}_{\mathrm{i}}+\mathrm{v}_{\mathrm{f}}}{2}\right)\Delta\mathrm{t}$

$=\left(\frac{-2+0}{2}\right)0,34$

$=-0,34\mathrm{~m}$

$Distance=0,34\mathrm{~m}$ (This part doesn't use Part 2)

$\Delta x=v_{i}\Delta t+1/2a\Delta t^2$

=(-2)(0,34)+1/2(5,8)(0,34)^2

$=-0,34\mathrm{~m}$

$\text{ Distance }=0,34\mathrm{m}$

$\Delta\mathrm{x}=\left(\frac{\mathrm{v}_{\mathrm{i}}+\mathrm{v}_{\mathrm{f}}}{2}\right)\Delta\mathrm{t}$

$=\left(\frac{-2+0}{2}\right)0,34$

$=-0,34\mathrm{~m}$

$Distance=0,34\mathrm{~m}$

There are more ways to answer this question. To view other options, click here.

4 marks

(4 x 1 mark)

Option 2

$\left.\begin{array}{l}W_{\text{net }}=\Delta E_{K}\\ F_{\text{net }}\Delta x\cos\theta=1/2m\left(v_{f}^2-v_{i}^2\right)\end{array}\right\}Anyone$

$(116)\Delta x\cos180^{\circ}=1/2(20)\left(0^2-2^2\right)$

$\Delta\mathrm{x}=0,34\mathrm{~m}$

Option 3

$\left.\begin{array}{l}W_{\text{net }}=\Delta E_{K}\\ W_{f}+W_{wll}=1/2mv_{f}^2-1/2mv_{i}^2\\ f\Delta x\cos\theta+\left(mg\sin30^{\circ}\right)\Delta x\cos\theta=1/2m\left(v_{f}^2-v_{i}^2\right)\end{array}\right\}Anyone$

$(18)\Delta x\cos180^{\circ}+(20)(9.8)\sin30^{\circ}\Delta x\cos180^{\circ}=y_2(20)\left(0^{\circ}-2^2\right)$

$\Delta\mathrm{x}=0,34\mathrm{~m}$

Option 4

$\left.\begin{array}{l}W_{\text{net }}=\Delta E_{K}\\ W_{f}+W_{w}=1/2mv_{f}^2-1/2mv_{i}^2\\ f\Delta x\cos\theta+mg\Delta x\cos120^{\circ}=1/2m\left(v_{f}^2-v_{i}^2\right)\end{array}\right\}Anyone$

$(18)\Delta x\cos180^{\circ}+(20)(9,8)\Delta x\cos120^{\circ}=1/2(20)\left(0^2-2^2\right)$

$\Delta x=0,34\mathrm{~m}$

Option 5

$\left.\begin{array}{l}W_{nc}=\Delta E_{p}+\Delta E_{k}\\ f\Delta x\cos\theta=mg\left(h_{f}-h_{i}\right)+1/2m\left(v_{f}^2-v_{i}^2\right)\end{array}\right\}Anyone$

$18\Delta x\cos180^{\circ}=20(9.8)\Delta x+1/2(20)\left(0^2-2^2\right)$

$-18\Delta x=196\Delta x\sin30^{\circ}-40$

$\Delta x=0,34\mathrm{~m}$