# Help with work and energy problem

• NAkid
In summary, the conversation discusses a problem involving a car traveling at a certain speed and distance and the force applied by the car on the road while stopping. The first part of the problem can be solved using work/energy concepts and the answer is 5402N. The second part involves finding the stopping distance when the car is traveling up a slope and the solution involves using a free body diagram and the formula Vf^2 = v0^2 + 2adx. However, the person is still having trouble getting the correct answer.
NAkid
Hi i need help with this problem. The first part asks
A car (m = 890.0 kg) traveling on a level road at 27.0 m/s (60.5 mph) can stop, locking its wheels, in a distance of 60.0 m (196.9 ft). Find the size of the horizontal force which the car applies on the road while stopping on the road. First solve this problem using work/energy concepts and then check your answer using kinematics/force law concepts.

I solved this by finding the acceleration and plugging into formula F=ma=5402N

Find the stopping distance of that same car when it is traveling up a 18.9deg slope, and it locks its wheels while traveling at 27.0 m/s (60.5 mph). Assume that muk does not depend on the speed.

I drew a free body diagram and came up with the following relationships
F-fk-mgsin(18.9)=ma where F=5402N
N-mgcos(18.9)=0

How do I solve for the horizontal distance? Is it just the vertical component of the Force?

I just tried using the formula Vf^2 = v0^2 + 2adx -- where final velocity=0, initial velocity is given, and i solved for a using the above equation F-ukmgcos(18.9)-mgsin(18.9)=ma. But I still didn't get the right answer..

For the first part of the problem, using work/energy concepts, we can determine the size of the horizontal force by using the equation W = Fd, where W is the work done, F is the force applied, and d is the distance traveled. In this case, the work done is equal to the change in kinetic energy, which can be calculated using the equation KE = 1/2mv^2. By solving for F, we get F = (m*v^2)/2d, which gives us a force of 5402 N.

To check our answer using kinematics/force law concepts, we can use the equation F = ma, where F is the net force, m is the mass, and a is the acceleration. We know the mass and the acceleration (which can be calculated using the formula a = v^2/2d), so we can solve for F, which again gives us a force of 5402 N.

For the second part of the problem, we can use the same equations to calculate the stopping distance on the slope. The only difference is that we now have to take into account the component of the force due to gravity acting on the car along the slope. This can be calculated using the equation Fg = mg*sinθ, where θ is the angle of the slope. We can then use this value to modify the equation for acceleration (a = v^2/2d) to include the gravitational force, giving us a = (v^2/2d) - (Fg/m). By solving for d, we can find the stopping distance on the slope. It is not just the vertical component of the force, as the force is acting at an angle and will contribute to both the horizontal and vertical components of acceleration.

## 1. How do I calculate work and energy in a problem?

To calculate work, you multiply the force applied by the distance over which the force acts. The formula is W = F * d. To calculate energy, you can use the formula E = m * g * h, where m is the mass of the object, g is the acceleration due to gravity, and h is the height of the object.

## 2. What is the difference between work and energy?

Work is the measure of the force applied to an object over a certain distance. Energy is the capacity of an object to do work. In other words, work is the action, while energy is the potential for action.

## 3. How do I know which units to use when calculating work and energy?

The units for work are joules (J) or newton-meters (N*m), while the units for energy are also joules (J). When calculating work, you will need to use units of force and distance, such as newtons (N) and meters (m). When calculating energy, you will need to use units of mass, acceleration, and distance, such as kilograms (kg), meters per second squared (m/s^2), and meters (m).

## 4. Can you provide an example of a work and energy problem?

Sure! Let's say you are pushing a box with a force of 50 newtons for a distance of 10 meters. To calculate the work done, you would use the formula W = F * d, so W = 50 N * 10 m = 500 J. Now, let's say you lift a 2 kg object up a flight of stairs that is 5 meters high. To calculate the energy used, you would use the formula E = m * g * h, so E = 2 kg * 9.8 m/s^2 * 5 m = 98 J.

## 5. What are some real-life applications of work and energy?

Work and energy are fundamental concepts in physics and can be seen in everyday life. Some examples include using energy to power our homes and vehicles, work being done when we lift or move objects, and energy being transformed from one form to another, such as in a roller coaster ride. Understanding work and energy can also help with understanding concepts like force, power, and mechanical advantage.

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