Exam review: don't remember how to do this

[Earth_kissed]

I have 2 problems I have forgot how to do.

1. An Arrow is shot at an angle of 45 degrees abouve the horizontal at 69 m/s. (a) How high will the arrow go? (b) what horizontal distance will the arrow travel.

these are the equations I know:

Dx=Vxt
(Dx: verticle distance, Vx: Verticle Velocity, t: time)

Dy=Vyt+1/2at^2
(Dy: Horizontal Distance, Vy: Horizontal Velocity, t: time, a: acceleration)

Vx=VIcosø
(Vx: Verticle Velocit, VI: I don't know what this means but it could mean initial velocity)

Vy=VIsinø

As far as this question goes... I have no clue... I think I was absent when my teacher went over this... but my exams are tomorrow and I don't have enough time time to ask him about it... please help.

2. What is the Coefficient of Friction for a 2000 kg car that skids to a stop with an acceleration of -3m/s^2

these are the equations I know for this problem... this problem isn't that hard but there doesn't seem to be enough information. I need a pushing force!

W=mg
(W:weight, M: mass, g: accel. due to gravity)

Fnet=ma
(Fnet: Net Force, a: acceleration)

Fnet=Fp+Ff
(Fp: pushing force, Ff: Frictional force)

Ff=µFn
(µ: coefficient of friction, Fn: Normal Force)

for this problem I know that the mass (2000 kg) and the acceleration (-3m/s^2) so Fnet=ma, Fnet= 2000*-3, Fnet=-6000.

I need to find Ff so I can solve for µ (Ff=µFn) because I already know Fn: W=mg, 2000*9.8, W=19600. and Fn=W when on a flat surface. so Fn=19600

to find Ff I'd use this equation: Fnet=Fp+Ff. I know Fnet=-6000 but I don't know Fp or Ff... Help

any help would be greatly apreciated!

P.S. I already know the answers (1.) 121m and 484m (2.) .306. all I need to know is how to get those answers!

Thanks

 

[Doc Al]

Originally posted by Earth_kissed
I have 2 problems I have forgot how to do.

1. An Arrow is shot at an angle of 45 degrees abouve the horizontal at 69 m/s. (a) How high will the arrow go? (b) what horizontal distance will the arrow travel.

Try this. Treat vertical and horizontal motion separately. (Find the components of the initial velocity in each direction.) First find the time it takes for the arrow to reach its highest point. (Use v_f = v_i + at, applied to the vertical motion.) Now find the height it reaches. (Use D_f=D_i+V_it+1/2at^2.) For part b, use D=V_it. (I'll let you figure out what t must be.)

2. What is the Coefficient of Friction for a 2000 kg car that skids to a stop with an acceleration of -3m/s^2

these are the equations I know for this problem... this problem isn't that hard but there doesn't seem to be enough information. I need a pushing force!

Who says you need a pushing force? The ground is doing all the pushing: The only horizontal force on the car is the friction of the ground against the tires. You have the acceleration, so find the force (F=ma). Then find the coefficient of friction: the normal force is the weight of car.

 

[M98Ranger]

for reaching out for help on the exam review. It sounds like you have a good understanding of the equations and concepts involved in these problems, but just need a refresher on how to apply them. Here are some steps to help you solve these problems:

1. For the first problem, you are given the initial velocity (69 m/s) and the angle (45 degrees) at which the arrow is shot. To find the maximum height, you can use the equation Dy=Vyt+1/2at^2. In this case, the initial velocity in the vertical direction (Vy) is equal to the initial velocity of the arrow (69 m/s) multiplied by the sine of the angle (45 degrees). So, Vy=69*sin(45) = 48.8 m/s. Since the arrow is shot at an angle, the acceleration due to gravity (a) will only affect the vertical component of the velocity. Therefore, you can use the value of a=-9.8 m/s^2 in the equation. You are trying to find the maximum height, so you can set Dy=0 and solve for t. This will give you the time it takes for the arrow to reach the maximum height. Once you have t, you can plug it back into the equation to find the maximum height (Dy). To find the horizontal distance traveled (Dx), you can use the equation Dx=Vxt. In this case, Vx=69*cos(45) = 48.8 m/s. You can use the same value for t that you found for the vertical component to find the horizontal distance traveled.

2. For the second problem, you are given the mass (2000 kg) and the acceleration (-3 m/s^2) of the car. You need to find the coefficient of friction (µ). To do this, you can use the equation Ff=µFn. In this case, Fn=mg, so Fn=2000*9.8=19600 N. You already know the value of Ff (from Fnet=ma, Fnet=-6000), so you can plug in these values and solve for µ. Ff=-6000 N and Fn=19600 N, so µ=Ff/Fn=-6000/19600=0.306.

I hope this helps you understand how to solve these problems. Remember to always check your units and use the correct equations for each situation