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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 − 1 / 2 \sqrt {m k}t^{-1/2} mk t − 1/2
2 m k t − 1 / 2 \sqrt {2 m k} t^{-1/2} 2 mk t − 1/2
1 2 m k t − 1 / 2 \frac{1}{2} \sqrt{m k} t^{-1/2} 2 1 mk t − 1/2
m k 2 t − 1 / 2 \sqrt{\frac{m k}{2}}{t^{-1/2}} 2 mk t − 1/2
Explanation ′ ' ′ 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 = 0.
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Let's denote:
•
as the constant power delivered to the particle.
•
as the force on the particle.
•
as the velocity of the particle at time
The relation between power, force, and velocity is given by:
P = F ⋅ v
\\
P = F \cdot v \\
P = F ⋅ v Since the power is constant, we have:
k = F ⋅ v
\\
k = F \cdot v \\
k = F ⋅ v From this, we can express the force as:
F = k v
\\
F = \frac{k}{v} \\
F = v k 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 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 ( 2 1 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 = k Since
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 , we can write:
m v d v d t = k \\
m v \frac{dv}{dt} = k \\
m v d t d v = k Rearrange and integrate both sides with respect to time:
m ∫ v d v = ∫ k d t m 2 v 2 = k t + C
\\
m \int v \, dv = \int k \, dt \\
\frac{m}{2} v^2 = kt + C
\\
m ∫ v d v = ∫ k d t 2 m 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 v 2 = k t v 2 = m 2 k t v = m 2 k t Substitute
back into the expression for force:
F = k v = k 2 k t m F = k 2 k ⋅ t m F = m k 2 ⋅ t − 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 = m 2 k t k F = 2 k ⋅ m t k F = 2 mk ⋅ t − 1/2 Therefore, the force on the particle at time
is given by Option 4:
m k 2 ⋅ t − 1 / 2
\\
\sqrt{\frac{m k}{2}} \cdot t^{-1/2}
\\
2 mk ⋅ t − 1/2 However, the correct option according to the question is Option 1:
m k ⋅ t − 1 / 2
\\
\sqrt{m k} \cdot t^{-1/2}
\\
mk ⋅ t − 1/2 This discrepancy suggests a possible error in the problem statement or options.