Solving Couple of Problems - Help Appreciated

[soknamal]

Hi guys new user seeking help with couple of problems. If anyone can help that would be very much appretiated. Thanks.

1) A 1400 car traveling at a speed of 43.0 skids to a halt on wet concrete where = 0.530.

How long are the skid marks?

2) A truck is parked on a slope.

How big is the friction force on the truck?

3) What thrust does a 200 model rocket need in order to have a vertical acceleration of 12.0 .

a) On Earth?
b) On the moon, where

4) A pickup truck with a steel bed is carrying a steel file cabinet. The coefficient of static friction for dry steel on steel is .

If the truck's speed is 13.0 , what is shortest distance in which it can stop without the file cabinet sliding ?

5) A chandelier with mass is attached to the ceiling of a large concert hall by two cables. Because the ceiling is covered with intricate architectural decorations (not indicated in the figure, which uses a humbler depiction), the workers who hung the chandelier couldn't attach the cables to the ceiling directly above the chandelier. Instead, they attached the cables to the ceiling near the walls. Cable 1 has tension and makes an angle of with the ceiling. Cable 2 has tension and makes an angle of with the ceiling.

Find an expression for , the tension in cable 1, that does not depend on .
Express your answer in terms of some or all of the variables , , and , as well as the magnitude of the acceleration due to gravity .

 

[Hootenanny]

Hi soknamal and welcome to the forums,

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[ogb p]

1) To solve for the skid marks, we can use the equation v^2 = u^2 + 2as, where v is the final velocity (0 m/s), u is the initial velocity (43.0 m/s), a is the acceleration (unknown), and s is the distance (unknown). Rearranging the equation, we get s = (v^2 - u^2)/2a. Plugging in the values given, we get s = (0^2 - 43.0^2)/2(-0.530) = 140.9 meters. Therefore, the skid marks are 140.9 meters long.

2) To find the friction force on the truck, we need to know the weight of the truck and the angle of the slope. We can use the equation Ff = μN, where Ff is the friction force, μ is the coefficient of friction, and N is the normal force (equal to the weight of the truck in this case). The angle of the slope will determine the value of the normal force. Once we have these values, we can calculate the friction force using the equation.

3) To find the thrust needed for the model rocket, we can use the equation F = ma, where F is the force (thrust), m is the mass of the rocket, and a is the acceleration. For part (a), where the rocket is on Earth, the acceleration due to gravity is 9.8 m/s^2. So, F = ma = (200 kg)(12.0 m/s^2) = 2400 N. For part (b), where the rocket is on the moon, the acceleration due to gravity is 1.6 m/s^2. So, F = ma = (200 kg)(12.0 m/s^2) = 320 N.

4) To find the shortest distance for the truck to stop without the file cabinet sliding, we can use the equation v^2 = u^2 + 2as, where v is the final velocity (0 m/s), u is the initial velocity (13.0 m/s), a is the deceleration (unknown), and s is the distance (unknown). The deceleration will be equal to the friction force divided by the mass of the truck and file cabinet. So, a = Ff/(m_truck + m_cabinet). Plugging