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(PHYS011)mt05-solutions.pdf
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Phys011/ 05 Mid-term Solution
1. This is a HW problem. The acceleration is reduced when the mass of the middle block is increased. The force on block 3, your back, is reduced.
2. (a) The work done W is given by the area under the force curve between x = 0.0 to x = 5.0 m. W = 2.0x2.0 C . x1.0x2.0 = 3.0 J. The work done is to increase the kinetic energy of the block, thus W = .K = . m (v2 C 2.02) = 3.0, v = (10)1/2 m/s.
(b)
Find the maximum acceleration a the smaller block will not slip.
The equation of motion for the smaller block is
fs C mg sin = ma,
where fs = n = mg cos,
a = g ( cos C sin) = g cos ( C tan).
The equation of motion for the larger block is
F C fs C Mg sin = Ma
F = mg cos + Mg sin + M g ( cos C sin)
= (m+M) g cos.
The answer for = 00 is F = (m+M)g which is what we get for the blocks on level ground. But when the incline angle exceed some value, the smaller block will slip so that the expression is not valid for = 900.
3. This problem is based worked example 6.4 and 6.5 (p.154-155 of Serway and Jewett) (a) The forces acting on the car are its weight mg, the normal reaction from the road n and the friction f.
The centripetal force is provided by the friction f
m(v2/R) = f = n = mg
v2 = Rg
v = (0.5x50x10)1/2 = 15.8 m/s
(b) The forces in this case are the weight and the normal force.
The horizontal component of the normal
force provides the centripetal force
n sin = m (v2/R).
The vertical component balances the weight
n cos = mg
thus v2 = gR tan, tan =
= 26.70
(c) The horizontal components of the normal force and friction provide the centripetal force n sin + f cos = m (v2/R)
n (sin + cos) = m (v2/R)
v is larger than in (b) due to the horizontal component of friction.
n cos = mg + f sin
f = n
n =mg / (cos C sin)
Finally
v2 = gR( sin + cos) / (cos C sin)
This expression reduces to
v2 = gR when = 0 (the case in (a))
and v2 = gR tan when = 0 (the case in (b)).
(d) There is a centripetal force to the left in a left turn. The frictional force from the seat and the contact force from the door provide the centripetal force.
4. (a) U(0.0) = 2.0 J, K(0) = 0.0. Since U(3.0) + K(3.0) = U(0.0) + K(0.0) = 2.0
K(3.0) = 2.0 C U(3.0)
K(3.0) = . (1.0) v2 = 2.0 C 1.0, v = (2.0)1/2.
F = C.U/.x,
F = 2.0 N between x = 0.0 to x = 1.0 m
F = 0.0 N between x = 1.0 to x = 2.0 m
F = C1.0 N between x = 2.0 to x = 4.0 m
The work done increases the kinetic energy. The work done W by F between x = 0.0 to x = 3.0 m is given by W = 2.0 C 1.0 = 1.0 J.
The increase in kinetic energy is . (1.0) (v2 C 0). Thus v = (2.0)1/2.
(b) To pull the object up the incline, one must overcome the combined forc