Need help on a question i need review for a quiz

  • Thread starter Thread starter albaboi123
  • Start date Start date
  • Tags Tags
    Quiz Review
AI Thread Summary
A 600-kg jet car accelerates from rest, traveling 1.00 km in 21.0 seconds before the jet engine shuts off, and then comes to rest after traveling an additional 1.4 km due to friction. To solve for the force of the jet engine and the constant frictional force, one can use kinematic equations to determine the car's acceleration during the first segment. The net force acting on the car can be calculated from this acceleration, taking into account the frictional force that stops the car afterward. Understanding the relationship between distance, time, and acceleration is crucial for solving the problem. This discussion emphasizes the application of kinematics and forces in physics.
albaboi123
Messages
5
Reaction score
0
i have this question that i couldn't get but it's on this quiz i have tommorow.

A 600-kg jet car accelerates from rest under the force of its jet engine.
After traveling 1.00 km in 21.0s, the jet engine shuts off and the jet car eventually comes to a rest at 1.4 km further down the track, stopped by the frictional force of the ground. Calculate the force due to the jet engine, and the constant frictional force applied while the car is in motion.

anyone can help me please?
 
Physics news on Phys.org
albaboi123 said:
i have this question that i couldn't get but it's on this quiz i have tommorow.

A 600-kg jet car accelerates from rest under the force of its jet engine.
After traveling 1.00 km in 21.0s, the jet engine shuts off and the jet car eventually comes to a rest at 1.4 km further down the track, stopped by the frictional force of the ground. Calculate the force due to the jet engine, and the constant frictional force applied while the car is in motion.

anyone can help me please?

(Jet) You know how far the car has gone (from rest) in 21 s and you know it has done so with a constant acceleration. Can you think of a Kinematics relation between x, t, and a?

(Friction) Again you can figure the constant acceleration, so you can figure the net force on the car. What is causing this net force?

-Dan
 
albaboi123 said:
i have this question that i couldn't get but it's on this quiz i have tommorow.

A 600-kg jet car accelerates from rest under the force of its jet engine.
After traveling 1.00 km in 21.0s, the jet engine shuts off and the jet car eventually comes to a rest at 1.4 km further down the track, stopped by the frictional force of the ground. Calculate the force due to the jet engine, and the constant frictional force applied while the car is in motion.

anyone can help me please?

(Jet) You know how far the car has gone (from rest) in 21 s and you know it has done so with a constant acceleration. Can you think of a Kinematics relation between x, t, and a?

(Friction) Again you can figure the constant acceleration (Hint: How fast is the car moving at 21 s?), so you can figure the net force on the car. What is causing this net force?

-Dan
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top