Ok so my professor posted a practice test and I need help on 4 questions from it

Question

Ok so my professor posted a practice test and I need help on 4 questions from it. Its supposed to not take that long to do but I can't understand how to get the correct answers. I have the correct answers for all of them except #3, so I just need someone to show me step by step how the answer the achieved. The correct answer is in bold and I didn't know how to post the pictures attached to the questions so its kinda bad quality. Also the equations below are some of the ones listed on the exam to use, thank you to whoever answers my questions!!

x = (1/2)at2 + vot             vav = x/t             a = (vf - vo)/t              vf2 = vo2 +2ax   

vav = (vf +vo)/2         SF = ma       S F = 0             f s= msh        f k= mkh     w = mg

Fc = mv2/r     SOHCAHTOA   

1.     Determine the tension in each of the ropes holding the object.

Ans: T1 = 1862N     T2 = 1620N

2.  Determine the tension in the ropes and the acceleration of the three blocks.

Ans:T1(between 16 and 8 kg blocks) = 52.32N

T2(between 3 and 12 kg blocks) = 78.48N

a = 3.27m/s^2

3.     Determine the force require to make a car (m=1000kg) turn a 25m radius curve at 23m/s. Is it likely the car could do this? Prove your answer.

Ans: I don't have the answer to this question.

4.     A rope applies a force T = 500N as shown as the 30kg block moves 14m along the incline. The length of the incline is unlimited. The coefficient of friction between the block and the plane is 0.33

a.     What is the acceleration of the block?

b.     If the force applied by the tension is removed after the block has traveled the 14m, how much farther will the block travel? Be specific, use numbers. Does the block slide back down the plane? Prove it.

Ans: a. The acceleration of the block while being pulled is 9.2m/s^2.

b. To determine distance traveled, you must calculate the speed of the block after the pull is finished (i.e. after the first 14m). This is 16m/s. Then you must calculate the acceleration of the block after the pull is removed. This is

Explanation / Answer

1. sum forces in the x

T2*cos(55) - T1 cos(60) = 0

T2 = T1 cos(60)/cos(55)

sum forces in the y

T2*sin(55) + T1*sin(60) -300*9.81 = 0

T1 cos(60)/cos(55)*sin(55) + T1*sin(60) -300*9.81 = 0

T1=1862

so T2 = 1862*cos(60)/cos(55) = 1620 N

2)

sum forces

12*9.81 = (12+8+16)*a
a = 3.27 m/s^2

for block 1
T1 = 16 a = 16*3.27 = 52.32 N

for block 2
T2 - T1 = 8 a
T2 = 52.32+8*3.27 = 78.48 N

3) F = m v^2/r = 1000*23^2/25= 21160 N
so unlikely

4. a) sum forces parallel

T - mg sintheta - friction = 0

sum force perp
N - m g cos theta = ma
N = mg cos theta

so T - m g sin tehta - u m g cos hteta = ma
500 - 30*9.81*sin(28 degrees) - 0.33*30*9.81*cos(28 degrees) = 30*a
a=9.2 m/s^2

b) first find speed
v^2 = v0^2 + 2 a x
v = sqrt(2*9.81*14)= 16.57 m/s


now need to find a with no tension
- 30*9.81*sin(28 degrees) - 0.33*30*9.81*cos(28 degrees) = 30*a
a=-7.46
now v^2 = v0^2 + 2 a x

0= 16.57^2 -2*7.46*x
x=18.4 m additional

c. yes since gravity> friction

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