October 10, 2007
PHY2053
Quiz 5 (5.4-5.7, 8.3)
Name:
UFID:
1. (5 pts) A 2.00 kg block is shot in the uphill direction from a light spring placed along a
frictionless incline of 45º. The spring has a spring constant of 350 N/m. The lower end of
the spring is fixed and the block is pressed against the upper end compressing the
spring by 8.00 cm from the equilibrium position. What is the speed the block when it
leaves the spring?
At the equilibrium position, the net force on the block is zero. Thus, the compression of
the spring at the equilibrium position (d) is given by
kd-mgsinθ = 0, d = mgsinθ/k = 2×9.8sin45º/350 = 0.0396 m
We compress the spring further by D = 0.08m and release. The block leaves the spring
when the spring comes back to its natural length. Since the work done by the spring and
gravity is converted to the kinetic energy, we get
KE = (1/2)k(d+D)²-mgsinθ(d+D) = 0.846J
(1/2)mv² = 0.843J, v = √(0.843×2/m) = 0.920 m/s
2. (5 pts) A 0.500 kg block is shot horizontally from a light spring having a spring
constant of 800 N/m. The block is pressed against the spring compressing the spring by
0.20 m and released. Then the block moves up a rough ramp with a coefficient of kinetic
friction of 0.30. The ramp is 1.00 m high and makes an angle of 30º to the horizontal.
Find the speed of the block after the block climbs up the ramp.
The kinetic energy of the block when it leaves the spring is
KEi = (1/2)kx² = 16.0J.
Some part of the energy is lost by the work done by friction and gravity, thus the kinetic
energy after the block climbs up the ramp is given by KEf = KEi-Wg-Wf, where
Wg = mgh = 0.5×9.8×1=4.90J and
Wf = μmgcosθ(1/sinθ) = 0.3×0.5×9.8×cos30×2 =2.55J
KEf = 16 – 4.9 – 2.55 = 8.55J
(1/2)mv² = 8.55J, v = √(2×8.55/0.5) = 5.85 m/s
3. (5 pts) A 600 kg elevator starts from rest on the ground and moves upward with a
constant acceleration for 5.00 s. At 5.00 s, the elevator is 10.0 m above the ground. Find
the average power of the elevator motor during this period.
Since P = Fv, we need calculate the tension exerted on the elevator and average velocity.
Δx = (1/2)at², a = 2Δx/t² = 2×10/5² = 0.8 m/s²
ma = T-mg, T = m(a+g) = 600(9.8+0.8) = 6360 N
v = Δx/Δt = 10/5 = 2.00 m/s
Thus, P = 6360×2 = 12720 w
4. (5 pts) A water molecule consists of an oxygen atom and two hydrogen atoms. The
distance between the oxygen atom and each of the hydrogen atoms is 0.1nm and two
bonds make an angle of 106º. Choose a coordinate system so that the oxygen atom is on
the positive y-axis and the two hydrogen atoms are on the x-axis and calculate the
center of the gravity of the molecule. An oxygen atom is 16 times as heavy as hydrogen.
All you have to do is express the positions of each atom in components and plug them
into the formula:
O = (0, 0.1cos53º) = (0 nm, 0.0602 nm)
H1 = (0.1sin53º, 0) = (0.0799 nm, 0 nm)
H2 = (-0.1sin53º, 0) = (-0.0799 nm, 0 nm)
Xcg = (m×0.0799-m×0.0799)/(18m) = 0 nm
Ycg = (16m×0.0602)/(18m) = 0.0535 nm