2 (Moderately Straightforward) Physics Questions on Mechanics & Kinematics

[Graffiti]

Question 1: If an object were thrown straight upward with an initial speed of 8 m/s, and it took 3 seconds to strike the ground, from what height was it thrown? (Ignore air resistance and take g = 10 m/s^2).

For this question, I assumed that it would take 1.5s to reach the peak hieght, so I plugged that into my equation.

d = (8)(1.5) + (1/2)(-10)(1.5)2
d =0.75 m

The answer doesn't really make sense.

Plugging in the time 3 seconds results in a more realistic answer (21m) but I'm confused as to when to divide time in half. I always thought you plug in the time it takes to reach the top, not the total time of flight. Am I doing something wrong here?

And second question: How do you normally approach Center of Mass questions. At first glance, they seem easy as heck, but after practicing, I was wrong.

Here's an example of what I'm having trouble with:

Question two: A uniform meter stick weighing 20 N has a 50-N weight on its left end and a 30-N weight on its right end. The bar is hung from a rope. What is the tension in the rope and how far from the left end of the bar should the rope be attached so that the stick remains level? (Answer: 100 N placed 40.0cm from the Left end of the bar).

I really don't know how to approach this problem. Answering the first part was easy, but given there's so many unknowns for the second portion of the question, its difficult for me to approach a solution.

FYI, both of these questions came from TPR Hyperlearning Book (Physics section).

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[Prateek]

Question 1: If an object were thrown straight upward with an initial speed of 8 m/s, and it took 3 seconds to strike the ground, from what height was it thrown? (Ignore air resistance and take g = 10 m/s^2).

For this question, I assumed that it would take 1.5s to reach the peak hieght, so I plugged that into my equation.

d = (8)(1.5) + (1/2)(-10)(1.5)2
d =0.75 m

The answer doesn't really make sense.

Plugging in the time 3 seconds results in a more realistic answer (21m) but I'm confused as to when to divide time in half. I always thought you plug in the time it takes to reach the top, not the total time of flight. Am I doing something wrong here?

And second question: How do you normally approach Center of Mass questions. At first glance, they seem easy as heck, but after practicing, I was wrong.

Here's an example of what I'm having trouble with:

Question two: A uniform meter stick weighing 20 N has a 50-N weight on its left end and a 30-N weight on its right end. The bar is hung from a rope. What is the tension in the rope and how far from the left end of the bar should the rope be attached so that the stick remains level? (Answer: 100 N placed 40.0cm from the Left end of the bar).

I really don't know how to approach this problem. Answering the first part was easy, but given there's so many unknowns for the second portion of the question, its difficult for me to approach a solution.

FYI, both of these questions came from TPR Hyperlearning Book (Physics section).

For #2:
Tension = Weight of all the things together = 20 + 50 + 30 = 100N

For the COM, pick any point on the stick as a reference, let's pick the very left end of the stick.

COM = [(50N)(0) + (30N)(1m) + (20N)(0.5m)] / [50N + 20N + 30N]

COM = [40 N*m] / [100N]
COM = 0.4m = 40cm

Remember to include the weight/center of mass of the stick as well (20N and 0.5m from the left)

 

[vsl5]

Question 1: If an object were thrown straight upward with an initial speed of 8 m/s, and it took 3 seconds to strike the ground, from what height was it thrown? (Ignore air resistance and take g = 10 m/s^2).

For this question, I assumed that it would take 1.5s to reach the peak hieght, so I plugged that into my equation.

d = (8)(1.5) + (1/2)(-10)(1.5)2
d =0.75 m

The answer doesn't really make sense.

Plugging in the time 3 seconds results in a more realistic answer (21m) but I'm confused as to when to divide time in half. I always thought you plug in the time it takes to reach the top, not the total time of flight. Am I doing something wrong here?

Do not assume the point at which he throws the ball is the ground, he might be standing on an elevated point or something and the ball goes all the way to the ground.
If Vi=8m/s, at the peak of the toss, V= 0. 0=8-10t t = 0.8s and distance traveled to the peak is 3.2m (0=8^2+2*g*d) so he has 3-0.8=2.2s of free fall. X=xo+1/2 at^2 x=0.5*10*2.2^2 ~20m
so Total distance traveled is approximately 23.2m (i rounded the 2.2^2=4)

And second question: How do you normally approach Center of Mass questions. At first glance, they seem easy as heck, but after practicing, I was wrong.

Here's an example of what I'm having trouble with:

Question two: A uniform meter stick weighing 20 N has a 50-N weight on its left end and a 30-N weight on its right end. The bar is hung from a rope. What is the tension in the rope and how far from the left end of the bar should the rope be attached so that the stick remains level? (Answer: 100 N placed 40.0cm from the Left end of the bar).

I really don't know how to approach this problem. Answering the first part was easy, but given there's so many unknowns for the second portion of the question, its difficult for me to approach a solution.

FYI, both of these questions came from TPR Hyperlearning Book (Physics section).

If in equilibrium, all forces must cancel. Fleft=Fright (useless in this case)
Fup=Fdown
T(CW)=T(CCW)

Fdown = 50+20+30N (weight of all components of the system
Fup = Tension in rope = T = 100N because Fdown=T as the system is not moving. Now set up the torque part of the question.

Take any point to be the point of rotation, I chose the left end because the question asked for how far from the left end I should attach the rope.
TCCW = 20N*0.5m + 30N*1m (0.5m because this is the center of mass of the board and 1m because it is a meter stick and the 30N weight all the way right).
T (CW) = Force from tension

T CW = T CCW
20*0.5 + 30*1 = 100N * x where x = distance from left where you attach rope. x=0.4m = 40cm

 

[collegestud2013]

Question 1: If an object were thrown straight upward with an initial speed of 8 m/s, and it took 3 seconds to strike the ground, from what height was it thrown? (Ignore air resistance and take g = 10 m/s^2).

They're basically asking for the displacement here, so use:

d = vi(t) + (1/2)at^2
and make sure to get your signs right (up is +, down is -), and to keep it simple just use d as the overall displacement of the entire flight and t as the total time. No need to find the max height or anything like some of the previous responses tried to do.

d = 8(3) + (1/2)(-10)(3^2) = 24 - 45 = -21 m

so the height would be 21 m

Question two: A uniform meter stick weighing 20 N has a 50-N weight on its left end and a 30-N weight on its right end. The bar is hung from a rope. What is the tension in the rope and how far from the left end of the bar should the rope be attached so that the stick remains level?

You know the tension in the rope will have equal the weight of all three things to be in equilibrium so to find that add up the weights: 50 N + 30 N + 20 N = 100 N
To makes things convenient use the left end of the meter stick as the pivot, and qualitatively you should see that the rope has to be closer to the 50 N weight. The torques have to sum to zero, and by convention i'm gonna use CCW as + and CW as -, so:

50(0) + (100)(x) - 20(.5) - 30(1) = 0
x = 40/100 = .4 m from left end