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Physics -- Determining Work -- Please help!!!!!!!!!!!!!!!!!!? |
1. A 900-N crate rests on the floor. How much work is required to move it at constant speed (a) 4.0 m along the floor against a friction force of 180 N, and (b) 4.0 m vertically? 2. A car does 7.0 x 10^4 J of work in traveling 2.8 km at constant speed. What was the average force of friction (from all sources) acting on the car? 3. A hammerhead with a mass of 2.0 kg is allowed to fall onto a nail from a height of 0.50 m . What is the maximum amount of work it could do on the nail? Why do people not just "let it fall" but add their own force to the hammer as is falls? 4. In pedaling a bicycle, a particular cyclist exerts a downward force of 90 N during each stroke. If the diameter of the circle traced by each pedial is 36 cm, calculate how much work is done in each stroke. Please help me do and understand these problems.. Any help is appreciated. Also, show your work so that I can understand what you are doing. Thanks a whole bunch! 1. (a) To move a 900 N crate at a constant speed across the floor means that the crate is not accelerating and the net force acting on it must be 0 N. Thus, the force applied to move the crate must be equal to the frictional force opposing the motion. Since friction is constant, the work done is simply: W = F * d = (180 N) * (4 m) = 720 N-m = 720 J (b) When lifting the crate vertically, there is no longer any friction. Again, to lift the crate at a constant speed means that the crate is not accelerating so the net force acting on it is 0 N. Thus, the force applied to raise the crate is equal to the crate's weight. Again, the work done is computed as: W = F * h = (900 N) * (4 m) = 3600 N-m = 3600 J 2. Again, assuming everything is constant, the work done by the car is equal to the force is exerts over a distance. Since its speed is constant, it's not accelerating so it's net force is zero, and the thus the force it exerts is equal to the forces (friction, etc) exerted on it. The equation, again, is: W = F * d, where W and d are given. Re-writing: F = W / d = (70,000 J) / (2.8 km * 1000 m/km) F = 25 N 3. Work is a form of energy; assuming, as your question states, that energy is conserved through the collision of the hammer and the nail, then the energy in the hammer before the collision is equal to the energy in the nail after the collision. This energy in the nail is equal to the maximum amount of work the nail can do. Using the conservation of energy, the energy of the hammer before the collision is: W = PE (potential energy of hammer) W = m * g * h W = (2 kg) * (9.8 m/s虏) * (.5 m) W = 9.8 J The hammer will be able to impart a maximum of 9.8 J of work to the nail. Notice that this is derived by simply letting the hammer fall; if we add our own forces, the energy in the hammer prior to striking the nail is increased. This allows more work to be done on the nail. 4. This problem is solved using the same reasoning as above: W = F * d In this case, we're given F and a pedal diameter. Notice that the pedal itself follows a circular path during the stroke, but this arc length is NOT the distance needed to compute the work. When computing work, the path that the mass takes is irrelevant; the net work only involves initial and final locations. In this case, the distance through which the force is exerted is in the downward direction only; this corresponds to the pedal diamter. Thus: W = F * d W = (90 N) * (.36 m) W = 32.4 J/stroke. Good Luck! :)if u sent me the equations u have to use i could do it...but i have forgotten the quations wow mama mia! |
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