CATAGORY: Mechanics
QUESTION: How much work is done by a person lifting a 2.0 kg object from
the bottom of a well at a constant speed of 2.0 m/s for 5.0 s (in kJ)?

Back to subcatalog.

The idea here is to utilize the formula P = F*v where P is the power output, F is the force, v is the velocity and * is the so called "dot" or "inner product" of the vectors F and v. In other words (1) P = |F||v|cos(a) Where a is the angle between F and v. In your case F is the force of gravity which is down and v is the velocity the object is being lifted which is up. The angle between F and v then is 180 degrees and so a = 180 degrees and cos(180) = -1.0. So with |F| = |mg| =2kg X 9.8m/s^2 and |v| = 2.0m/s then P = 2kg X 9.8m/s^2 X 2.0m/s X (-1.0) = -39.2 J/s Note: the negative means that energy is being expended. Now the person lifts for a total of 5.0s so the total energy EXPENDED is E = P X time = 39.2 J/s X 5.0s = 196 J *********************** * * * E = 196 J * * * ***********************

CATAGORY: Mechanics QUESTION: Identicle particles are placed at the 50-cm and 80-cm marks on a meter stick of negligible mass. This rigid body is then mounted so as to rotate freely about a pivot at the 0-cm mark on the meter stick. If this body is released from rest in a horizontal position, what is the angular speed of the meter stick as it swings through its lowest position (in rad/s)? This is a good conservation of energy with rotation problem. Lets call the zero of potential energy when the stick is at its horizontal position. Since the stick is released from that point then the rotational kinetic energy is initialy zero as well. With conservation of total energy we have (1) initial total energy = final total energy The initial total energy is zero however so (2) 0 = final total energy The final total energy is the potential energy (PE) of each particle plus the rotational kinetic energy (KE) of each particle thus (3) 0 = PE(1) + PE(2) + KE(1) + KE(2) If particle 1 is at 50cm (=.5m) and particle 2 is at 80cm (=.8m) then (4) PE(1) = -Mg.5m PE(2) = -Mg.8m Notice the minus sign. This is because the particles are BELOW the starting point which we have taken to be zero PE ---> 0 height level. The rotational KE of a particle of mass M at radius r is (5) KE = .5Mr^2 w^2 (note: ^2 means "squared") Where w is the angular rotation rate in radians per unit time. Thus (6) KE(1) = .5M(.5m)^2 w^2 KE(2) = .5M(.8m)^2 w^2 M is the same for both expressions and since both are on the same rigid stick then w is the same for both too. Substituting (4) and (6) into (3) gives (7) 0 = -Mg.5m -Mg.8m + .5M(.5m)^2 w^2 + .5M(.8m)^2 w^2 Dividing through by M gives (8) 0 = -g.5m - g.8m + .5(.5m)^2 w^2 + .5(.8m)^2 w^2 or taking -g.5m - g.8m =-(1.3m)g to the other side of the equation and factoring w^2 gives us (9) (1.3m)g = [.5(.5m)^2 + .5(.8m)^2]w^2 (1.3m)g = (.445m^2) w^2 w^2 = (1.3m)g/(.445m^2) =(1.3m)(9.8m/s^2)/(.445m^2) = 28.6 s^-2 w = sqrt(28.6 s^-2) ~ 5.4 rad/s ****************** * * * w ~ 5.4 rad/s * * * ******************

Back to subcatalog.