A small but dense 2.0-kg stone is attached to one end of a very light rod that i
ID: 1416300 • Letter: A
Question
A small but dense 2.0-kg stone is attached to one end of a very light rod that is 1.2 m long. The other end of the rod is attached to a frictionless pivot. The rod is raised until it is vertical, with the stone above the pivot. The rod is released and the stone moves in a vertical circle with no air resistance. What is the tension in the rod as the stone moves through the bottom of the circle?
20 N
40 N
60 N
80 N
100 N
0.00 W
1.3 W
2.5 W
2.2 W
5.0 W
20 N
40 N
60 N
80 N
100 N
The net force that an animal exerts on a large fruit it has found is observed over a 10-s interval and is shown in the graph in the figure. What was the average power delivered to the fruit by the animal over this time interval?
0.00 W
1.3 W
2.5 W
2.2 W
5.0 W
Explanation / Answer
We need to find the velocity of the stone at the bottom
We can use the conservation of energy to find the velocity at the bottom
Before the stone starts rotating the energy of the stone is its potential energy = m g h
Where h is the height from the bottom h = 2 x 1.2 = 2.4 m
Ei = m g h
As the stone reaches the bottom its potential energy is completely converted to the kinetic energy . The total energy at that position is
Ef = (1/2) m v2
Equating the total energies
Ei = Ef
m g h = (1/2) m v2
v = sqrt (2 g h )
v = sqrt (2 x 9.8 x 2.4)
v = 6.86 m /s
Thus the velocity of the stone at the bottom is 6.86 m/s
The tension in the rod is
T = m v2 / r + Fg
Where m v2 / r is the centripetal force and Fg is the gravitational force
T = m v2 / r + m g
m = 2.0 kg , r = 1.2 m
T = 2 x (6.86)2 / 1.2 + (2 x 9.8)
T = 98.03 N = 100 N
B) From the graph the work done by the force can be found by calculating the area under the curve
Area of the triangle is A = (1/2) x 10 x 5 = 25
Thus the work done is W = 25 J
The power is P = W / t = 25 /10 = 2.5 W