Physics : Force and Motion Questions - Please explain work

[Krackers300]

I am having difficulty understanding the explanation of the answers in the EK book. Would someone be able to explain how they they worked out the answers to the following questions please ?

Thanks !

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[AdaptPrep]

Hi, Krakers300--

For Item #1, it is easiest to break it into two right triangles.

Triangle 1:
Ignore the height of the balloon at first. The two vectors to be added are:
From point A to point B...10 km (north)
From point B to point C...8 km (east)
The sum (use Pythagorean theorem) would be from point A to point C...square root of 164 km. Keep it as square root of 164 for now.

Triangle 2 (where we now consider the height):
From point A to point C...square root of 164 km
From point C to point D (the height)...6 km
The sum (use Pythagorean theorem again) would be from point A to point D...square root of 200 km which is about 14 km. Therefore, the answer would be C (14 km)

I hope that helps.

 

[AdaptPrep]

For Item #14:

Since the rock was thrown horizontally, it will only be in the air for as long as it falls 40 m (the height of the cliff). The horizontal speed will tell us how far from the base of the cliff the rock will fall (which is not asked). Remember that projectile motion has a x-component and a y-component. Only the y-component is used to determine the amount of time the object is in the air. Since the rock was thrown horizontally, it has no y-component initial velocity.
Using the kinematic equation:
d = (vi)(t) + (1/2)(a)(t^2)
Where d is distance of y-component (-40 m), vi is initial velocity in the y-component (0 m/s), t is time, a is rate of acceleration in y-component (-10 m/s^2):
-40 = (0)(t) + (1/2)(-10)(t^2)
-40 = -(5)(t^2)
40 = (5)(t^2)
8 = t^2
t = 8^(1/2) seconds = 2.8 seconds. Therefore, the correct answer is A (3 seconds).

I hope that makes better sense.

 

[AdaptPrep]

For Item #17:

There are two methods to approach this type of problem.

If you like using sin, cos, and tan, use method 1:
Sketch the right triangle where the height is 20 m and the hypotenuse is 40 m. The angle above the horizontal (theta) is therefore 30 degrees (sin theta = 20/40). The rate of acceleration down a frictionless ramp of an angle theta above the horizontal is: g sin theta (where g is rate of acceleration due to gravity in free fall). Thus, the rate of acceleration down the ramp is 10 sin 30 = 5 m/s^2.
Now, use the kinematic equation:
d = (vi)(t) + (1/2)(a)(t^2)
Where d is distance (40 m), vi is initial velocity (0 m/s), t is time, a is rate of acceleration (5 m/s^2):
40 = (0)(t) + (1/2)(5)(t^2)
40 = (2.5)(t^2)
16 = t^2
t = 4 seconds. Therefore, the correct answer is C (4 seconds).

If you don't like using sin, cos, and tan, use method 2:
Calculate how long it would take to free-fall from its height of 20 m by using the kinematic equation:

d = (vi)(t) + (1/2)(a)(t^2)
Where d is distance (-20 m), vi is initial velocity (0 m/s), t is time, a is rate of acceleration (-10 m/s^2):
-20 = (0)(t) + (1/2)(-10)(t^2)
-20 = -(5)(t^2)
20 = (5)(t^2)
4 = t^2
t = 2 seconds
Next, set up a ratio:
distance of ramp/distance of free-fall = time down ramp/time of free-fall
40/20 = t/2
t = 4 seconds (option C)
This works due to similar triangles and ratios (it is where "g sin theta" from method 1 is derived as well).

I hope one of those methods makes sense for you.

Good Luck!

​

 

[AdaptPrep]

For Item #20:

Hooke's Law for a spring is:
F = -k delta y
The added mass of 0.5 kg stretched the spring 1 cm (downward is -1 cm). F also equals mass times rate of acceleration due to gravity (F = ma = mg):
F = -k delta y
mg = -k delta y
(0.5 kg)(10 m/s^2) = -k (-1 cm)
5 N = k (1 cm)
5 N/cm = k

You could have also done this by unit conversion. The answer had to be in N/cm. The difference in mass was 5 N (0.5 kg * 10 m/s^2) and the distance stretched was 1 cm. 5 N / 1 cm = 5 N/cm.

Happy studying!

 

[AdaptPrep]

For Item #23:

We will use the kinematic equation in order to calculate the acceleration of the box upwards:
vf = vi + at
where vf is final velocity (5 m/s), vi is initial velocity (0 m/s), a is rate of acceleration, and t is time (1 s).
5 m/s = 0 m/s + a (1s)
a = 5 m/s^2

So the box has an acceleration of 5 m/s^2 upward. Now, the free body diagram will have a force acting upward (Fu; a positive value) and force downward due to gravity or weight (Fw; a negative value). The sum of these two forces is Fu - Fw = ma.
The upward force Fu is what we need to solve for:
Fu = ma + Fw = ma + mg
Fu = (10kg)(5 m/s^2) + (10 kg)(10 m/s^2)
Fu = 50 N + 100 N
Fu = 150 N

So I think the answer is A (150 N).

I hope that helps.