A confusing word problem. Can any one solve it?

In summary, the toy car can climb a 24° incline at 16m/s and can travel on a horizontal table at 39m/s. The car has a kinetic energy of 0.003072 j and a potential energy of 0.1531 j. The car can climb a 24° incline with a speed of 28m/s.
  • #1
katiesstaar
6
0

Homework Statement



A 240gram toy car is driven by an electric motor that has a constant power output. The car can climb a 24° incline at 16m/s and can travel on a horizontal table at 39m/s. The frictional force retarding the motion is kv, where k is a constant and v is its speed. How steep an incline can it climb with a speed of 28m/s?


The Attempt at a Solution



I've tried solving this for 2 days now and but not succeeded. It seems quite simple but it is the kv part that I don't know how to use.
Please could someone explain how to solve it?

Help is very much appreciated!
Thank you!
 
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  • #2
katiesstaar said:

Homework Statement



A 240gram toy car is driven by an electric motor that has a constant power output. The car can climb a 24° incline at 16m/s and can travel on a horizontal table at 39m/s. The frictional force retarding the motion is kv, where k is a constant and v is its speed. How steep an incline can it climb with a speed of 28m/s?


The Attempt at a Solution



I've tried solving this for 2 days now and but not succeeded. It seems quite simple but it is the kv part that I don't know how to use.
Please could someone explain how to solve it?

Help is very much appreciated!
Thank you!

Welcome to the PF.

The k*v part is just like wind resistance and/or rolling resistance. They are saying that the only retarding force on flat ground is this resistance, which is proportional to velocity.

Do the 3 FBDs for the 3 cases, and include this resistant force in each. You should end up with a set of equations that you can use to solve for the velocity in the 3rd situation. Please show us your work...
 
  • #3
Oh thank you :)
I still seem to be going in circles.

I made a table with the kinetic and potential energies for each one: (by the way the speeds are supposed to be in cm/s - i typed it wrong)

1) 0.39m/s --> Kinetic E = m(v^2)/2 = 0.018252 J
Potential E = mgh = 0 J

2) Kinetic = 0.003072 J
Potential = 0.1531 J

3) Kinetic = 0.009408 J
Potential = ?? (i don't know theta, that is what i need to find)

So i guess 2) is best to work with since i have both values so i can find friction:
so friction is PE - KE = 0.1531 - 0.003072 = 0.150028 J

I have no idea if this is right...
and how do i include the "motor has a constant power output" ?

0.39m/s
 
  • #4
katiesstaar said:
so friction is PE - KE = 0.1531 - 0.003072 = 0.150028 J

I have no idea if this is right...
You are going about this wrong by trying to use energy concepts and incorrect equations. You need to draw free body diagrams as berkeman has suggested, and apply Newton's first law.
and how do i include the "motor has a constant power output" ?
Power is force times velocity, that is, P = Fv. The power delivered by the motor is the same for all cases. Write down all givens and relevant equations. Please show your work using free body diagrams.
 
  • #5


I understand the frustration of encountering a confusing word problem. In order to solve this problem, we need to use the principles of physics, specifically Newton's laws of motion.

First, let's define our variables. We know that the toy car has a mass of 240 grams, and the electric motor has a constant power output. The car can climb a 24° incline at 16m/s and can travel on a horizontal table at 39m/s. The frictional force retarding the motion is kv, where k is a constant and v is its speed.

To solve for the maximum incline the car can climb at a speed of 28m/s, we need to use the equation for work done against gravity, which is W = mgh, where m is the mass of the car, g is the acceleration due to gravity (9.8 m/s^2), and h is the height of the incline.

We also need to take into account the work done against friction, which is W = Fd, where F is the frictional force and d is the distance traveled.

Since we are given the speed (v) and the frictional force (kv), we can use the equation for kinetic energy, which is KE = 1/2mv^2, to find the value of k.

Once we have the value of k, we can plug it into the equation for work done against friction and solve for the maximum incline (h) at a speed of 28m/s.

I hope this explanation helps you solve the problem. Remember to always define your variables and use the appropriate equations to solve for the unknowns. Good luck!
 

1. What is the word problem asking for?

The word problem is asking for a solution or answer to a question posed in a mathematical or logical format.

2. What makes the word problem confusing?

The word problem may be confusing due to the use of unfamiliar or complex language, multiple steps or variables, lack of context or background information, or a combination of these factors.

3. How can I approach solving a confusing word problem?

There are several strategies that can be used to solve a confusing word problem, such as breaking it down into smaller parts, drawing diagrams or illustrations, using trial and error, or creating an equation or formula to represent the problem.

4. What should I do if I am still unable to solve the word problem?

If you are still unable to solve the word problem, it may be helpful to seek assistance from a teacher, tutor, or classmate. You can also try looking for similar problems online or in a math textbook for guidance.

5. How can I avoid getting confused by word problems in the future?

To avoid getting confused by word problems in the future, it may be helpful to practice regularly with different types of problems, read the problem carefully and identify key information, and ask for clarification if needed. It may also be beneficial to understand and practice different problem-solving strategies.

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