# Physics question help (1 Viewer)

#### Farhanthestudent005

##### Member

I need help with this question. Will appreciate if someone helped and briefly explains it. Thank you!

#### Attachments

• 31.4 KB Views: 5

#### cossine

##### Member
I believe num 2 mg*sin(theta) up the slope as force parallel to the slope is mg*sin(theta).

The force normal to the slope is mg*cos(theta).

Make sure to double check. Have not done physics in a while.

#### yashbb

##### Member
its mgcostheta up the slope cause its opposing direction of movement.

#### ExtremelyBoredUser

##### Well-Known Member
mgsintheta up the slope. Normal force opposes weight force and they balance each other out on an inclined plane. Frictional force applies to driving force on the mass so it would be up the slope. Since the mass is stationary, net force is 0 so all forces cancel each other out, hence frictional force is mgsintheta in the opposite direction/up the slope.

#### A1La5

##### Member
its mgcostheta up the slope cause its opposing direction of movement.
$\bg_white mgcos(\theta)$ would act in the direction that is perpendicular to the slope, rather than up or down the slope.

As the net force is equal to zero (Object is at rest -> Net force is zero as a result of Newton's First Law of Motion), the horizontal (parallel to the inclined plane) and vertical (perpendicular to the inclined plane) forces are equal to each other. Using the above diagram as a reference:

$\bg_white mgcos(\theta) = N$

and

$\bg_white mgsin(\theta) = f$

#### icycledough

##### Member
Well, I don't think we can actually answer the question without the graph which is mentioned in the question. Is there any way you could send that through as well?

#### Farhanthestudent005

##### Member
Well, I don't think we can actually answer the question without the graph which is mentioned in the question. Is there any way you could send that through as well?
There you go

#### Lith_30

##### New Member
For part a:
momentum is calculated by $\bg_white p=mv$ therefore momentum would be...
$\bg_white \\p=1500\times20\\\\p=30000kgms^{-1}$

For part b:
We have the values for initial velocity, final velocity and displacement and we need to calculate acceleration, therefore the most suitable equation would be $\bg_white v^2=u^2+2as$, rearrange to get $\bg_white a=\frac{v^2-u^2}{2s}$
$\bg_white \\a=\frac{0^2-20^2}{2(1300)}\\a=-0.15384....\\a=-0.15ms^{-2}$

Since the average acceleration is negative, this means that the car is accelerating the opposite direction which is south. Therefore average acceleration is $\bg_white 0.15ms^{-2}\text{South}$

Last edited:

#### OreoMcFlurry

##### New Member

I would say that you should tick only the second box (ie. The hovercraft reached a speed of approximately 30m/s)

This is the graph that we were given:

As you may remember, Impulse is the change in momentum:

This is basically the area under the graph, which is (30 x 10)/2 {ie. Area of triangle}
Therefore, Impulse = 150 kgm/s

As mentioned before, Impulse is the change in momentum (v is final velocity and u is initial velocity):
150 = mv - mu = (5 x v) - (5 x 0)

Therefore, 150 = 5v
and so, v=30m/s

Now, the other options shouldn't be selected as:
• The acceleration changes over time - as the force experienced by the hovercraft changes with respect to time and F=ma (where m is 5kg), therefore the acceleration is not constant, the acceleration is only 3m/s^2 when t=10 seconds.

• The hovercraft didn't increase in KE by 150 J

KE actually increases by 2250 J

• The hovercraft didn't travel 5m, if you use this kinematics equation:

The distance travelled by the hovercraft is actually 150m.

#### Attachments

• 38.8 KB Views: 2
• 1.5 KB Views: 2

#### Farhanthestudent005

##### Member

I can try part c on my own. All I need for this is just part a and b. Please somebody help urgently. I need this for my upcoming physics exam tomorrow. Thanks