Two forces of magnitude 4 N and 8 N are acting on box when the box moves rightwards across a frictionless table. the speed of the box at time t is 1 m/s. What is the change in kinetic energy of the box?
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The initial K. E. of the box=0
The rightwards force acting on the box F=m(dv/dt)=4 cosθ+8 cosϕ
or m∫dv= (4 cosθ+8 cosϕ)∫ dt
or mv=(4 cosθ+8 cosϕ) t+c
Now at time t the velocity v=1m/s so
c=m-(4 cosθ+8 cosϕ)t
Therefore the above eqn becomes; mv= (4cosθ+8 cosϕ)t+m-(4 cosθ+8 cosϕ)t=m
or v=1
Therefore the the K. E. at time t becomes ; 1/2mv2 =1/2m × 1 × 1= m/2
So the change in K. E.=Final K. E.-Initial K. E. =m/2-0= m/2
The rightwards force acting on the box F=m(dv/dt)=4 cosθ+8 cosϕ
or m∫dv= (4 cosθ+8 cosϕ)∫ dt
or mv=(4 cosθ+8 cosϕ) t+c
Now at time t the velocity v=1m/s so
c=m-(4 cosθ+8 cosϕ)t
Therefore the above eqn becomes; mv= (4cosθ+8 cosϕ)t+m-(4 cosθ+8 cosϕ)t=m
or v=1
Therefore the the K. E. at time t becomes ; 1/2mv2 =1/2m × 1 × 1= m/2
So the change in K. E.=Final K. E.-Initial K. E. =m/2-0= m/2
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