Вопрос
2. A person weighing 60 kg, standing on ice skates, throws a load weighing 3 kg in the hori- zontal direction at a speed of 8m/s relative to the ice The coefficient of friction between the skates and the ice is 0 .02. a) What is the initial impulse of the load? b) What is the speed of a person immediately after the throw? b) How far will a person travel after throwing to a stop?
Решения
4.4
(206 Голоса)
Илларион
мастер · Репетитор 5 лет
Ответ
a) The initial impulse of the load can be calculated using the formula:Impulse = Change in momentumThe initial momentum of the load is zero since it is at rest. After the throw, the momentum of the load is given by:Momentum = mass x velocitySubstituting the given values, we have:Momentum = 3 kg x 8 m/s = 24 kg·m/sTherefore, the initial impulse of the load is 24 kg·m/s.b) To find the speed of the person immediately after the throw, we can use the conservation of momentum. The total momentum before the throw is equal to the total momentum after the throw.Let's denote the velocity of the person after the throw as v. The total momentum before the throw is zero since both the person and the load are at rest. After the throw, the momentum of the person is given by:Momentum = mass x velocitySubstituting the given values, we have:Momentum = 60 kg x vThe total momentum after the throw is the sum of the momentum of the person and the momentum of the load:Total momentum = Momentum of person + Momentum of load0 = 60 kg x v + 24 kg·m/sSolving for v, we get:v = -24 kg·m/s / 60 kg = -0.4 m/sTherefore, the speed of the person immediately after the throw is 0.4 m/s.c) To find the distance traveled by the person after throwing to a stop, we can use the work-energy theorem. The work done by the friction force will be equal to the change in kinetic energy of the person.The work done by the friction force is given by:Work = Friction force x DistanceThe friction force can be calculated using the formula:Friction force = Coefficient of friction x Normal forceThe normal force is equal to the weight of the person, which is given by:Normal force = mass x acceleration due to gravitySubstituting the given values, we have:Normal force = 60 kg x 9.8 m/s^2 = 588 NFriction force = 0.02 x 588 N = 11.76 NNow, we can calculate the work done by the friction force:Work = 11.76 N x DistanceThe change in kinetic energy of the person is given by:Change in kinetic energy = Final kinetic energy - Initial kinetic energyThe initial kinetic energy of the person is zero since they are at rest. After the throw, the kinetic energy of the person is given by:Kinetic energy = 1/2 x mass x velocity^2Substituting the given values, we have:Kinetic energy = 1/2 x 60 kg x (0.4 m/s)^2 = 4.8 JTherefore, the work done by the friction force is equal to the change in kinetic energy:11.76 N x Distance = 4.8 JSolving for Distance, we get:Distance = 4.8 J / 11.76 N = 0.408 mTherefore, the person will travel approximately 0.408 meters after throwing to a stop.