A 510-kg roller coaster car starts at a height of 32.0 m. (A kinematic equation could be also used to find the initial velocity.)Ĭalculating Work | Analysis of Situations Involving External Forcesģ1. Substituting and solving for velocity (v) yields v = 27.2 m/s. The initial velocity of the car can be found using the KE equation: KE = 0.5*m*v 2 where m=878 kg and KE i=323632 J. Substituting and solving yields a = -8.16 m/s/s.ĭ. The friction force is the net force (since the up and down forces balance) and the mass is 878 kg. The acceleration of the car can be found using Newton's second law of motion: F net = m*a Rearrange the equation and it takes the form KE i = -W nc. (PE i and PE f = 0 J since the car is on the ground and KE f = 0 J since the car is finally stopped). The initial kinetic energy can be found using the work-energy theorem. Substituting and solving yields -323632 J (rounded to -324000 J).ī. Where F=7160 N, d=45.2 m, and Theta=180 degrees (the force is in the opposite direction as the displacement). The work done upon the car can be found using the equation W = F*d*cos(Theta) The average force acting upon the car is 7160 N. An 878-kg car skids to a stop across a horizontal surface over a distance of 45.2 m. Potential Energy | Kinetic Energy | Definition and Mathematics of Work | Powerģ0. The power of the hiker can be found by dividing the work by the time. (PE i = 0 J since the hiker starts on the ground and KE i = KE f since the speed is constant these two terms can be dropped from the equation since they are equal). The work done upon the hiker can be found using the work-energy theorem. The final potential energy is 21888 J and the initial potential energy is 0 J. The potential energy change can be found by subtracting the initial PE (0 J) from the final PE (m*g*h f). The speed of the hiker is constant so there is no change in kinetic energy - 0 J.ī. the potential energy change of the hiker.Ī.If it takes 384 s to climb the hill, then determine. A 51.7-kg hiker ascends a 43.2-meter high hill at a constant speed of 1.20 m/s. Now divide the work by the time to determine the power: P = W/t = (1019.2 J)/(1.20 s) = 849 Wattsĭefinition and Mathematics of Work | PowerĢ9. Where F = m*g = 637 N (the weight of the 65.0 kg object), d =1.60 m and Theta = 0 degrees (the angle between the upward force and the upward displacement). The work done in elevating his 65.0-kg mass up the stairs is determined using the equation W = F*d*cos(Theta) PSYWĮddy's power is found by dividing the work which he does by the time in which he does it. Eddy, whose mass is 65.0-kg, climbs up the 1.60-meter high stairs in 1.20 s.

Thus, W=Delta PE = m*g*(delta h) = (3.00 kg)*(9.8 m/s/s)*(0.45 m) = 13.2 Jĭefinition and Mathematics of Work | Potential EnergyĢ8. The work done equals the potential energy change. The second method is to recognize that the work done in pulling the cart along the incline at constant speed changes the potential energy of the cart.

(The angle theta represents the angle betwee the force and the displacement vector since the force is applied parallel to the incline, the angle is zero.) Substituting and solving yields W = F*d*cos(Theta) = (20.8 N)*(0.636 m)*cos(0) = 13.2 J. The first method involves using the equation W = F*d*cos(Theta)

There are two methods of solving this problem. Determine the work done upon the cart and the subsequent potential energy change of the cart. A force of 20.8 N is applied parallel to the incline to lift a 3.00-kg loaded cart to a height of 0.450 m along an incline which is 0.636-m long. A student applies a force to a cart to pull it up an inclined plane at constant speed during a physics lab. Substituting these values into the above equation yields W = F*d*cos(Theta) = (24.5 N)*(6 m)*cos(0) = ~150 J (147 J)Ģ7. In this case, the d=6.0 m the F=24.5 N (it takes 24.5 N of force to lift a 2.5-kg object that's the weight of the object), and the angle between F and d (Theta) is 0 degrees. The work done upon an object is found with the equation W = F*d*cos(Theta) Approximate the work required lift a 2.5-kg object to a height of 6.0 meters. Part B: Straightforward Computational ProblemsĢ6.

The Review Session » Work and Energy Packet » Answers Q#26-36 Work and Energy Review

Coast rider desent angle