Three marbles are released at the same height of 0.40 m on a 'frictionless' metal ramp. They then roll across a carpeted floor.
The red marble (m=0.500kg) rolls to a stop in Region B.
Sketch the scenario, labeling important points on the sketch.
Using EPCs, indicate the types of energy present at each important point.
Predict where the green and blueish marbles are going to stop.
A motorcycle and large transport truck are traveling at the same speed on a busy highway. Ahead is a traffic jam and both drivers need to stop their vehicles in 100 meters or they will crash into the back of the stopped cars ahead of them. They both slam on their brakes at the same time and SKID to a stop. Suppose the tires of the motorcycle and truck are made of the same material and both are traveling at the same velocity.
The motorcycle stops just before hitting the cars, predict where the truck would stop.
For your assigned scenario, complete the following on a whiteboard:
Sketch of scenario
List of items within your system
Using arrows, indicate the direction and distance the force was applied to the system
Draw a LoL diagram showing the showing the work done on the system.
Energy Pie Charts at significant points in the motion of the object.
Compare the energies of the ball at the bottom of each ramp.
As the Ferrari 488 is coming to a stop, a force of friction is acting to slow the car to a stop. Suppose a car is traveling at 112.7 km/h (31.3 m/s). With today's modern antilock braking systems, the 1470-kg car is able to stop in 39.6 m. The brakes apply an incredible constant 18 100 N force throughout the braking process.
Determine the velocity of the $284,700 sports car at these distances after braking:
10 m
19.8 m
30 m
39.6 m
Spreadsheet with solution.
For more information regarding the Energy Bar Charts discussed in class, here are some resources:
Internal Energy => ∆Therm
Determine the work done on the lawn mower by the person in if he exerts a constant force of 75.0N at an angle 35º below the horizontal and pushes the mower 25.0m on level ground?
The equation for the work is
𝑊=𝐹𝑑cos𝜃
Substituting the known values gives
𝑊 = (75.0 N)(25.0 m)cos(35.0º)
W= (75.0 N)(25.0 m)cos(35.0º)
1536 J=1.54×10^3 J.
Adapted from: Open Stax College Physics
2. Convert the amount of work from joules to kilocalories and compare it with the person mowing the lawn average daily intake of 10,000kJ (about 2400kcal of food energy. One calorie (1 cal) of heat is the amount required to warm 1 g of water by 1ºC, and is equivalent to 4.186J, while one food calorie (1 kcal) is equivalent to 4186J.
Converting the work in joules to kilocalories yields
𝑊=(1536J)(1kcal/4186J)=0.367kcal
The ratio of the work done to the daily consumption is
𝑊 / (2400kcal) = 1.53×10 ^-4
Please make a copy of the following document: Energy Transfer and Power
Submit a copy of your work on the Google Classroom assignment: Energy Transfer and Power
This also listed on 02.3f - Power