Download PHY 2048: Physic 1, Discussion Section 3885 Quiz 5

TA: Tomoyuki Nakayama
Monday, October 4, 2010
PHY 2048: Physic 1, Discussion Section 3885
Quiz 5 (Homework Set #6)
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The only force acting on a 0.800 kg body as the body moves along an x axis varies as shown in the
figure below right. The scale of the figures’ vertical axis is set by Fs = 4.00 N. The velocity of the
body at x = 0 is 6.00 m/s.
a) At what value of x will the body have a kinetic
energy of 10.0 J?
The kinetic energy of the body at x = 0 is
Ki = (1/2)mvi2 = 14.4 J
In the region between x = 0 and 2 m, the force does
some positive work and then some negative work. By
symmetry, the total work is zero in this region. (Recall
that work done by a force is given by the area under
the force vs position graph, with signs included.) This
means the kinetic energy increases first and then
decreases to the original value. Thus the kinetic energy
of the body is always larger than 14.4 J. In the region
between x = 2 m and 5 m, the kinetic energy of the body is expresses as
K = 14.4 + (-Fs)(x - 2)
Setting the energy to be 10 J, and we solve the equation for x.
10 = 14.4 – Fs(x – 2) ⇒ x = (14.4 – 10)/Fs + 2 = 3.1 m
b) What is the maximum kinetic energy of the body between x = 0 and x = 5.00 m?
The kinetic energy of the body increases between x = 0 and 1 m and then decreases. Therefore, the
kinetic energy takes the maximum value at x = 1 m. The maximum energy is
Kmax = K (x = 1 m) = Ki + (1/2)Fs×1 = 16.4 J