Answer
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Hint: For the given question first, try to figure out what is speed and work. Speed is the distance covered by an object per time. It is also defined as the rate at which work is done. The relationship between speed and work can be obtained by using the work and kinetic energy theorem.
Complete answer:
Speed is the distance traveled by a body per time taken to cover that distance.
$v = \dfrac{d}{t}$ , where v is the speed d is the distance and t is the time taken.
Work is defined as the energy transferred while moving an object over a distance with the application of force. It is the product of force and distance.
$W = F \times d$ , where f is the force and d is the distance covered.
Kinetic energy is the energy possessed during the motion of an object.
$K = \dfrac{1}{2}m{v^2}$ , where K is for kinetic energy, m is the mass and v is the speed.
From kinematics, we know the relationship between initial speed \[\left( u \right)\] , final speed \[\left( v \right)\] and distance \[\left( d \right)\] of an object is
${v^2} = {u^2} + 2ad$
Multiplying both sides $\dfrac{m}{2}$ , we get
$ \Rightarrow $$\dfrac{1}{2}m{v^2} = \dfrac{1}{2}m{u^2} + mad$
$ \Rightarrow $$\dfrac{1}{2}m{v^2} - \dfrac{1}{2}m{u^2} = mad$
$ \Rightarrow $$\dfrac{1}{2}m{v^2} - \dfrac{1}{2}m{u^2} = F \times d$ ($F = ma$, From Newton’s second law}
$ \Rightarrow {K_2} - {K_1} = W$
$ \Rightarrow \Delta K = W$
Kinetic energy is proportional to the square of speed. As speed increases kinetic energy also increases which will eventually give positive work. (When \[u < v\] , \[W\] will be positive)
As speed decreases kinetic energy also decreases which in turn gives a negative work. (When\[u > v\] , \[W\] will be negative)
Note: First try to understand the definition of work, speed, and kinetic energy. Derive the work – kinetic energy theorem and figure out how the change in speed affects work.
Complete answer:
Speed is the distance traveled by a body per time taken to cover that distance.
$v = \dfrac{d}{t}$ , where v is the speed d is the distance and t is the time taken.
Work is defined as the energy transferred while moving an object over a distance with the application of force. It is the product of force and distance.
$W = F \times d$ , where f is the force and d is the distance covered.
Kinetic energy is the energy possessed during the motion of an object.
$K = \dfrac{1}{2}m{v^2}$ , where K is for kinetic energy, m is the mass and v is the speed.
From kinematics, we know the relationship between initial speed \[\left( u \right)\] , final speed \[\left( v \right)\] and distance \[\left( d \right)\] of an object is
${v^2} = {u^2} + 2ad$
Multiplying both sides $\dfrac{m}{2}$ , we get
$ \Rightarrow $$\dfrac{1}{2}m{v^2} = \dfrac{1}{2}m{u^2} + mad$
$ \Rightarrow $$\dfrac{1}{2}m{v^2} - \dfrac{1}{2}m{u^2} = mad$
$ \Rightarrow $$\dfrac{1}{2}m{v^2} - \dfrac{1}{2}m{u^2} = F \times d$ ($F = ma$, From Newton’s second law}
$ \Rightarrow {K_2} - {K_1} = W$
$ \Rightarrow \Delta K = W$
Kinetic energy is proportional to the square of speed. As speed increases kinetic energy also increases which will eventually give positive work. (When \[u < v\] , \[W\] will be positive)
As speed decreases kinetic energy also decreases which in turn gives a negative work. (When\[u > v\] , \[W\] will be negative)
Note: First try to understand the definition of work, speed, and kinetic energy. Derive the work – kinetic energy theorem and figure out how the change in speed affects work.
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