Physics 225 F’03 – Dr. Shapiro
Midterm Exam #2 (Section 3)
Closed book. One 3”´5” note card, the handout and a calculator may be used. Please do not write on the test or the handout. Both must be turned in with your test answers. Note: This is one hour exam.
Partial credit will be given. Please show all your work.
1) In the diagram shown below the angle between the incline plane and the horizontal is adjustable. The coefficient of kinetic friction mk between the block and the incline plane is 0.15. Initially, a block with mass 5 kg rests on the incline plane a distance 1 m up the plane, and the angle that the incline plane makes with the horizontal is 15 deg.
(a) If the block is just on the verge of slipping, what is the coefficient of static friction?
At the point the block
is just about to slip, the downhill component
of the gravitational force is just balanced by the uphill frictional
force. So 
so 
(b) The incline plane angle is now suddenly increased to 16 deg, and the block slides down the plane. What is the net force on the block?
Now the net downhill
force is the gravitational component along the
incline plane minus the force of kinetic friction uphill.
So 
, where a is taken to be positive
downhill. So F = 6.45N downhill.
(c) How long does it take for the block to reach the bottom of the incline plane?
From part (b) the a
= F/m = 1.29m/s2, so
from 
we have 
so t = 1.245s.
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2) The first hill on a roller coaster ride is 50 m above ground level. The first valley is 5 m above ground level. The second hill is 40 m above ground level. The mass of the roller coaster cars and the passengers totals 1500 kg. The cars have a speed of 0.1 m/s when they go over the first hill. Ignore friction both in the wheel bearings and from air resistance, and assume that the mass of the wheels is negligible compared to the total mass of the cars and passengers.
(a) What is the change in gravitational potential energy from the first hill to the second hill?
, note the sign is important.
(b) What is the speed of the cars as they go over the second hill?
The loss
of potential energy is balanced by a gain in kinetic energy. So, 
, but the first term on the right is very small
(only 7.5J)
so we can ignore it and v2 = 14m/s.
3) A 2 kg block is pressed against a spring with a spring constant of 80 N/m until the spring is compressed 0.2 m from its equilibrium position. As shown in the diagram below, the surface immediately under the spring and block is frictionless. However, after the block is released it eventually encounters a surface where mk = 0.15.
(a) When the spring is compressed, how much energy is stored in it?
From 
we find that the
stored energy in the spring is 1.6J.
(b) When the block is released (but before it reaches the surface with friction), what is its kinetic energy?
When the block is released all the potential energy stored in the spring is converted to kinetic energy of the block so K = 1.6J.
(c) How far does the block travel on the surface with friction before it stops?
The block will stop
when the work done by the frictional force
equals the kinetic energy the block had when it first encountered the
surface
with friction. So, 
and d = 0.544m.
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4) A 1200 kg cannon fires an 50 kg shell horizontally with a speed of 550 m/s relative to the muzzle. The cannon is located on a frictionless surface, so it can recoil freely.
(a) What is the speed of the shell with respect to the ground?
Let m1
refer to the mass of the cannon, and m2 refer to the
mass of
the shell. Before the shell is fired
the momentum of the system is zero, so after the shell is fired we have
where v1
and v2 are velocities with respect to the
ground. But 
so 
which implies that 
(b) What is the kinetic energy of the shell with respect to the ground?
(c) What is the speed of the cannon with respect to the ground?
From 
so the speed is just 22.0
m/s.
(d) What is the kinetic energy of the cannon with respect to the ground?
