1.
In the figure, a frictionless roller coaster car of mass m = 827 kg tops the first hill
with speed vo = 17.0 m/s at height h = 50.3 m. If the gravitational energy of the
car-earth system is taken to be zero at B, what is its value when the car is at C?
Give the answer in kJ and give three significant figures.
ΔU = UC-UB=mg(-h/2)
With UB = 0
Uc = -mgh/2 = -827*9.8*50.3/2 = - 203831 J = -204 kJ
2.
A 0.72 kg ball thrown directly upward with an initial speed of 14 m/s reaches a
maximum height of 7.9 m. What is the change in the mechanical energy of the ballEarth system during the ascent of the ball to reach that maximum height? Give the
answer in J and give two significant figures. (Do not assume conservation of
mechanical energy in this problem, use the given data to find ΔEmech )
ΔEmech = ΔK+ΔU =(1/2)m(0-v2)+mgh =-0.5*0.72*142 + 0.72*9.8*7.9 = -15 J
3.
A 553 g block is released from rest at height ho above a vertical spring with spring
constant k = 537 N/m and negligible mass. The block sticks to the spring and
momentarily stops after compressing the spring 19.0 cm. What is the value of ho?
Give the answer in cm and give three significant figures.
W = ΔK+ΔU+ΔEth = ΔK + ΔUs + ΔUg +ΔEth
For the system block+earth, there is no external force, so W = 0. Also ΔEth = 0 - no
non-conservative forces in the system.
This reduces to conservation of mechanical energy:
ΔK + ΔU = 0
ΔK + ΔUs + ΔUg = 0
(0-0) + (1/2)k(x2-0) - mg(h+x) = 0
(1/2)*537*0.1902 – 0.553*9.8*(h+0.190) = 0
Î h = 1.60 m = 160 cm
4.
A pendulum consists of a 6.5 kg stone swinging on a 4.0 m string of negligible
mass. The stone has a speed of 7.5 m/s when it passes through its lowest point.
What is the speed when the string is at 60o to the vertical? Give the answer in m/s
and give two significant figures.
W = ΔK+ΔU+ΔEth = ΔK + ΔUs + ΔUg +ΔEth
For the system stone+earth, the only external force is the tension from the string
which does no work. Therefore W = 0. Ignoring air resistance ΔEth = 0.
Î ΔK + ΔU = 0
(1/2)m(vf2 - vi2)+ mgh = 0
(1/2)*m*( vf2 – vi2) + m*g*(L -L Cosθ) = 0
(1/2)*m*( vf2 – 7.52) + m*9.8*4.0*(1-Cos[60°]) = 0
Solving for vf Æ vf = 4.1 m/s
5.
In the figure, a block of mass m = 12 kg is released from rest on frictionless incline
of angle θ=30o. Below the block is a spring that can be compressed 2.0 cm by a
force of 270 N. The block momentarily stops when it compresses the spring by 4.6
cm. How far does the block move down the incline from its rest position to this
stopping point? Give the answer in m and give two significant figures.
W = ΔK+ΔU+ΔEth = ΔK + ΔUs + ΔUg +ΔEth
Initial Æ when the block is released from rest
Final Æ when the block stops after compressing the spring by x=4.6 cm -maximum compression – block at rest momentarily.
For the system block+incline+earth, there is no external force; therefore W = 0.
Frictionless incline Æ ΔEth = 0.
ΔK + ΔUs + ΔUg = 0
(0-0) + (1/2)k(x2-0) - mg(d*Sinθ) = 0
In the above equation x = 4.6 cm, m = 12 kg, g =9.8 m/s2, θ = 30° and d is the
unknown that needs to be found. However k is not directly given in the question, it
has to be found applying Hooke’s law as: k = F/x = 270/0.02 = 13500 N/m.
(1/2)*13500*0.0462 – 12*9.8*d*Sin[30] = 0
Î d = 0.24 m
6.
A rope is used to pull a 3.84 kg block at constant speed 5.72 m along a horizontal
floor. The force on the block from the rope is 7.67 N and directed 13.7o above the
horizontal. What is the increase in the thermal energy (= ΔEth = -Wf ) ? (Use
conservation of energy principle to answer this question, so you may learn how to
use this principle). Give the answer in J and give three significant figures.
W = ΔK+ΔU+ΔEth
For the system block+floor+earth, there IS external force therefore W ≠ 0. But DU
= 0 as there is no change in the gravitational potential energy (block is moving
horizontally and there is no springs. ΔK is also zero because block moves with a
constant speed.
ΔEth = W
= F·d
= F d Cos θ
= 7.67*5.72*Cos[13.7]
= 42.6 J
7.
You push a 1.8 kg block against a horizontal spring, compressing the spring by 15
cm. Then you release the block, and the spring sends it sliding across a tabletop. It
stops 75 cm from where you released it. The spring constant is 367 N/m. What is
the block-table coefficient of kinetic friction? Give two significant figures.
W = ΔK+ΔU+ΔEth = ΔK + ΔUs + ΔUg +ΔEth
Initial Æ when the spring is compressed to x=15 cm
Final Æ when the block stops after moving d=75 cm.
For the system block+floor+earth, there is no external force from the chosen initial
to final; therefore W = 0. ΔUg = 0 as the block moves horizontally.
0 = ΔK + ΔUs +ΔEth
0 = (0-0) + (1/2)k(0 - x2) + μk N d
0 = -(1/2)kx2 + μk mg d (N = mg by applying Newton’s 2nd law on the block in the
vertical direction)
μk = kx2 / (2 mg d)
= (367*0.152) / (2*1.8*9.8*0.75)
= 0.31
8.
Two blocks, of masses M = 2.5 kg and 2M, are connected to a spring of spring
constant k=390 N/m that has one end fixed, as shown in the figure. The horizontal
surface and the pulley are frictionless, and the pulley has negligible mass. The
blocks are released from rest with the spring relaxed. What maximum distance does
the hanging block fall before momentarily stopping? Give the answer in m and give
two significant figures.
W = ΔK+ΔU+ΔEth = ΔK + ΔUs + ΔUg +ΔEth
Initial Æ when the spring is relaxed and blocks are released from rest
Final Æ when the spring is stretched to the maximum, again blocks are at rest.
Spring is stretched to an unknown length x and both blocks moves by the
same amount, x)
For the system block+table+earth, there is no external force from the chosen initial
to final; therefore W = 0. Frictionless surface Æ ΔEth = 0
0 = ΔK + ΔUs + ΔUg
0 = (0-0) + (1/2)k(x2 – 0) + (2M) g(- x)
(1/2)kx2=2Mg x
Î x = 4Mg / k = 4*2.5*9.8/390 = 0.25 m
(only the mass 2M makes a change in
the gravitational potential energy)