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A particle of mass m m m is driven by a machine that delivers a constant power k k k watt. If the particle starts from rest, the force on the particle at time t t t is
hard
Work, Energy and Power
2015
physics
m k t ā 1 / 2 \sqrt {m k}t^{-1/2} mk ā t ā 1/2
2 m k t ā 1 / 2 \sqrt {2 m k} t^{-1/2} 2 mk ā t ā 1/2
1 2 m k t ā 1 / 2 \frac{1}{2} \sqrt{m k} t^{-1/2} 2 1 ā mk ā t ā 1/2
m k 2 t ā 1 / 2 \sqrt{\frac{m k}{2}}{t^{-1/2}} 2 mk ā ā t ā 1/2
Explanation ā² ' ā² To solve this problem, we need to find the force on a particle given that a machine delivers constant power k k k watt. \\
The particle starts from rest, so its initial velocity v 0 = 0. v_0 = 0.
\\
v 0 ā = 0.
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Let's denote:
ā¢
as the constant power delivered to the particle.
ā¢
as the force on the particle.
ā¢
as the velocity of the particle at time
The relation between power, force, and velocity is given by:
P = F ā
v
\\
P = F \cdot v \\
P = F ā
v Since the power is constant, we have:
k = F ā
v
\\
k = F \cdot v \\
k = F ā
v From this, we can express the force as:
F = k v
\\
F = \frac{k}{v} \\
F = v k ā Next, we need to find the expression for velocity
as a function of time
The kinetic energy
of the particle is given by:
K = 1 2 m v 2
\\
K = \frac{1}{2} m v^2
\\
K = 2 1 ā m v 2 The rate of change of kinetic energy is equal to the power:
d d t ( 1 2 m v 2 ) = k
\\
\frac{d}{dt}\left(\frac{1}{2} m v^2\right) = k
\\
d t d ā ( 2 1 ā m v 2 ) = k Differentiating the kinetic energy with respect to time, we get:
m v d v d t = k
\\
m v \frac{dv}{dt} = k \\
m v d t d v ā = k Since
d v d t \frac{dv}{dt} d t d v ā is the acceleration
we have:
m v a = k \\
m v a = k \\
m v a = k Using
a = d v d t , a = \frac{dv}{dt}, a = d t d v ā , we can write:
m v d v d t = k \\
m v \frac{dv}{dt} = k \\
m v d t d v ā = k Rearrange and integrate both sides with respect to time:
m ā« v ā d v = ā« k ā d t m 2 v 2 = k t + C
\\
m \int v \, dv = \int k \, dt \\
\frac{m}{2} v^2 = kt + C
\\
m ā« v d v = ā« k d t 2 m ā v 2 = k t + C Since the particle starts from rest, at
t = 0 , v = 0 , t = 0, v = 0,
t = 0 , v = 0 , so
Thus, we have:
m 2 v 2 = k t v 2 = 2 k t m v = 2 k t m
\\
\frac{m}{2} v^2 = kt \\
v^2 = \frac{2kt}{m}
\\
v = \sqrt{\frac{2kt}{m}}
\\
2 m ā v 2 = k t v 2 = m 2 k t ā v = m 2 k t ā ā Substitute
back into the expression for force:
F = k v = k 2 k t m F = k 2 k ā
t m F = m k 2 ā
t ā 1 / 2
\\ F = \frac{k}{v} = \frac{k}{\sqrt{\frac{2kt}{m}}}
\\ F = \frac{k}{\sqrt{2k} \cdot \sqrt{\frac{t}{m}}}
\\ F = \sqrt{\frac{m k}{2}} \cdot t^{-1/2}
\\
F = v k ā = m 2 k t ā ā k ā F = 2 k ā ā
m t ā ā k ā F = 2 mk ā ā ā
t ā 1/2 Therefore, the force on the particle at time
is given by Option 4:
m k 2 ā
t ā 1 / 2
\\
\sqrt{\frac{m k}{2}} \cdot t^{-1/2}
\\
2 mk ā ā ā
t ā 1/2 However, the correct option according to the question is Option 1:
m k ā
t ā 1 / 2
\\
\sqrt{m k} \cdot t^{-1/2}
\\
mk ā ā
t ā 1/2 This discrepancy suggests a possible error in the problem statement or options.