Problem 6: A 0.50-kg block attached to an ideal spring with a spring constant of
ID: 1512575 • Letter: P
Question
Problem 6: A 0.50-kg block attached to an ideal spring with a spring constant of 8ON/m oscillates ona horizontal frictionless surface. When the spring is 4.0 cm longer than its equilibrium length, the speed of the block is 0.50m/s. The greatest speed of the block is: A. 0.23m/s B. 0.32m/s C. 0.55m/s D. 0.71m/s E. 0.93m/s Problem 7: Two spacemen are floating together with zero speed in a gravity-free region of space. The mass of spaceman A is 120 kg and that of spaceman B is 90 kg. Spaceman A pushes B away from him with B attaining a final speed of 0.5m/s. The final recoil speed of A is: A. zero B. 0.38m/s C. 0.5m/s D. 0.67m/s E. 1.0m/s Problem 8: A 0.2-kg rubber ball is dropped from the window of a building. It strikes the sidewalk below at 30m/s and rebounds up at 20m/s. The impulse on the ball during the collision is: A. 10N s upward B. 10N s downward C. 2.0N s upward D. 2.0N s downward E. 9.8N s upwardExplanation / Answer
6) Option D) is correct.
Spring Energy = 1/2KX^2
Kinetic Energy = 1/2mv^2
At the time given, our initial time for our initial energy, it has some kinetic energy and some spring energy.
At the time of it's greatest speed, our final time or final energy, it will be at equilibrium, x=0 and will have all kinetic energy. Since Energy Initial = Energy Final we can set spring energy + kinetic energy at 4 cm = kinetic energy at 0 cm.
Therefore,
1/2 (80) (0.04)^2 +1/2 (0.5) (0.5)^2 = 1/2 (0.5) (v)^2 solve for v and you have your answer also the x needs to be in m not cm
0.064 + 0.0625 = (1/4) v^2
v^2= 0.506
v= 0.71 m/s
7) Option B) is correct.
We start this and all momenta problems by noting the total momentum before the event. In this case, as neither space dude is moving, as best as we can tell, the total momentum is P = 0. Therefore, after the push off, the total momentum must also be zero re the conservation of momentum law.
Thus we have Mv + mV = 0 and M and m are the two masses, with v = ? and V = 0.5 m/s as the velocities.
Solve for v = -(m/M)V = - (90/120)*0.5 = -0.375 m/s.
Where the negative sign means in the direction opposite to the direction of the V = 0.5 of B.
8) Impulse = change in momentum = m*dv = 0.20*(vfinal - vinitial) = 0.2[+20 - (-30)] = 10 N.s
5) Option B) is correct.
Work = force * distance
Since the surface is level, friction force = µk * m * g
µk = coefficient of kinetic friction
m * g = weight
friction force = 0.05 * 5000 = 250 N
Since the sled is moving at a constant speed, the acceleration = 0 m/s^2. So the net force on the sled = 0 N. Since the friction force is in the opposite direction of the motion of the sled, Net force = Force pulling – Friction force.
Force pulling – 250 = 0
Force pulling = 250 N
Work = 250 N * 1000 m = 250,000 N *m = 2.5*10^5 J