Kinematics/Force help - especially with direction

[xcfzm94]

Homework Statement

1. An object has a mass of 45kg and is pulled along a horizontal surface by a rope that makes an angle 32 above the horizontal. If the coefficient of friction for the surface is 0.13 and the tension on the rope is 95N what is the acceleration of the object along the horizontal? (0.66m/s is the ans in my textbook)

2. A student is standing 10m from a tall building, if he throws a ball at 12m/s at an angle of 35 up from the horizontal at what height will the ball hit the bulidng (1.92m)

3. [10kg]----[20kg]----> 100N (applied) [weights in boxes]
Given the diagram find the following if the surface has a coefficient of friction of 0.30

Find the accleration of the system (2.06m/s) i keep getting 0.39
Find the tension connecting the 10kg block to the 20kg block (50N)

4. If a student throws and object into the air with a speed of 16m/s at an angle of 25 above the horizontal, if the building is 75m tall how far from the base will the object hit the ground

5. A young woman swims at 2m/s@ 45 S of W for 5 minutes then she swims at 1.5m/s @ 80 E of N for 4 min
What is the average velocity of the woman (40.94 m/min @ 79 S of W ) i can't seem to find this answer

It may seem like alot, but I've tried my best at these questions and can't figure them out please even one will help

@phantom what you see is exactly what i see i can take a picture if you really need the [10kg]--[20kg]---- are boxes being pulled on a rope by 100N applied force

Homework Equations

f= ma
v=d/t
a=v/t
d=vit+1/2at^2
Ff=μ|fg|

The Attempt at a Solution

1. no idea
2. no idea

3. fg+fg+t=ma
a = fg+fg+t/m
fg1 = 10 * 9.81 *.3 = 29.43
fg2= 20 * 9.81 *.3 = 58.86

100N + (-29.43N) + (-58.86N)/30kg = 0.39m/s
Tension if i cannot get accerlation right i can't get the tension right so i didnt even try.

4. no idea
5. 2m/s * 600s = 1200m
1.5 * 240s= 360

(-1200 * cos45 + 360 * cos10)^2 + (-1200*sin45 + 360 * sin10)^2 - than square rooted didnt get the right answer

 

[PhanthomJay]

You must show what your attempt was so that we may better assist in determining where you are stuck.

Also, posting 5 problems in one post is a bit much. It s best to post each problem separately, showing your attempt, in order for you to get adequate help for each.

You might want to start with problem 3, showing how you arrived at your answer, and better describing the problem as given. Neither your answer nor the book's make sense without a better description and wording of the problem.

 

[PhanthomJay]

Problem 3: if the coefficient of kinetic friction is 0.3 for each block, your answer for the acceleration is correct. I don't know where the book answer of 2.06 comes from.

 

[Xisune]

Question 5:

You made a mistake in the math, 5 minutes is 300 seconds, not 600 seconds.

 

[Nickie]

Hello,

I am happy to assist you with your questions on kinematics and force. To begin, it is important to understand the basic concepts and equations involved in these topics.

Kinematics refers to the study of motion, specifically the position, velocity, and acceleration of objects. Force, on the other hand, is a push or pull on an object that can cause it to accelerate. In order to solve problems involving kinematics and force, we need to use equations such as Newton's Second Law (f=ma) and the equations of motion (v=d/t, a=v/t, d=vit+1/2at^2).

Now, let's go through each of the questions and break them down:

1. In this question, we are given the mass of the object (45kg), the angle of the rope (32 degrees), the coefficient of friction (0.13), and the tension on the rope (95N). To find the acceleration of the object, we can use Newton's Second Law: f=ma. The force acting on the object is the tension on the rope (95N) minus the force of friction (μ|fg|). The force of friction can be calculated by multiplying the coefficient of friction (0.13) by the normal force (which is equal to the weight of the object, so μ|fg|=0.13*45*9.81=58.8675N). Therefore, the net force acting on the object is 95N-58.8675N=36.1325N. Now, we can plug this into Newton's Second Law: 36.1325N=45kg*a. Solving for a, we get a=36.1325N/45kg=0.8025m/s^2. This is the acceleration of the object along the horizontal.

2. In this question, we are given the initial velocity of the ball (12m/s), the angle of the throw (35 degrees), and the height of the building (10m). We can use the equations of motion to solve for the height at which the ball will hit the building. First, we need to find the time it takes for the ball to reach the building. We can use the equation v=d/t, where v is the vertical component of the initial velocity (12sin35=6.88m/s) and d is the height of the building (10m). Sol