NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXI06:WORK ENERGY AND POWER
355694
Work done in time \(t\) on a body of mass \(m\) which is acceleration from rest to speed \(v\) in time \(t_{1}\) as a function of time \(t\) is given by:
1 \(\dfrac{1}{2} m v^{2}\left(\dfrac{t}{t_{1}}\right)^{2}\)
\(a=v / t_{1}\), Displacement after time \(t\): \(S=\dfrac{1}{2} a t^{2}=\dfrac{1}{2} \dfrac{v}{t_{1}} t^{2}\) \(W=F S=m a S=m \dfrac{v}{t_{1}} \dfrac{1}{2} \dfrac{v}{t_{1}} t^{2}=\dfrac{1}{2} m v^{2}\left(\dfrac{t}{t_{1}}\right)^{2}\)
PHXI06:WORK ENERGY AND POWER
355695
A truck accelerates from speed \(v\) to 2 \(v\). Work done in during this is
1 Three times as the work done in accelerating it from rest to \(v\)
2 Same as the work done in accelerating it from rest to \(v\)
3 Four times as the work done in accelerating it from rest to \(v\)
4 Lless than the work done in accelerating it from rest to \(v\)
Explanation:
Let \(m\) be mass of the truck. According to work-energy theorem \(W = {K_f} - {K_i}\) Work done in accelerating the truck from rest to speed \(v\) is \(\begin{aligned}& W_{1}=\dfrac{1}{2} m v^{2}-0 \quad(\because u=0) \\& =\dfrac{1}{2} m v^{2}\end{aligned}\) Work done in accelerating the truck from speed \(v\) to 2\(v\) is \(W_{2}=\dfrac{1}{2} m(2 v)^{2}-\dfrac{1}{2} m(v)^{2}\) \(=\dfrac{3}{2} m v^{2}=3 \dfrac{1}{2} m v^{2}=3 W_{1} \quad(\operatorname{Using}(1))\)
KCET - 2013
PHXI06:WORK ENERGY AND POWER
355696
A pendulum of length 2 \(m\) left at \(A\). When it reaches \(B\), it looses \(10 \%\) of its total energy due to air resistance. The velocity at \(B\) is
355694
Work done in time \(t\) on a body of mass \(m\) which is acceleration from rest to speed \(v\) in time \(t_{1}\) as a function of time \(t\) is given by:
1 \(\dfrac{1}{2} m v^{2}\left(\dfrac{t}{t_{1}}\right)^{2}\)
\(a=v / t_{1}\), Displacement after time \(t\): \(S=\dfrac{1}{2} a t^{2}=\dfrac{1}{2} \dfrac{v}{t_{1}} t^{2}\) \(W=F S=m a S=m \dfrac{v}{t_{1}} \dfrac{1}{2} \dfrac{v}{t_{1}} t^{2}=\dfrac{1}{2} m v^{2}\left(\dfrac{t}{t_{1}}\right)^{2}\)
PHXI06:WORK ENERGY AND POWER
355695
A truck accelerates from speed \(v\) to 2 \(v\). Work done in during this is
1 Three times as the work done in accelerating it from rest to \(v\)
2 Same as the work done in accelerating it from rest to \(v\)
3 Four times as the work done in accelerating it from rest to \(v\)
4 Lless than the work done in accelerating it from rest to \(v\)
Explanation:
Let \(m\) be mass of the truck. According to work-energy theorem \(W = {K_f} - {K_i}\) Work done in accelerating the truck from rest to speed \(v\) is \(\begin{aligned}& W_{1}=\dfrac{1}{2} m v^{2}-0 \quad(\because u=0) \\& =\dfrac{1}{2} m v^{2}\end{aligned}\) Work done in accelerating the truck from speed \(v\) to 2\(v\) is \(W_{2}=\dfrac{1}{2} m(2 v)^{2}-\dfrac{1}{2} m(v)^{2}\) \(=\dfrac{3}{2} m v^{2}=3 \dfrac{1}{2} m v^{2}=3 W_{1} \quad(\operatorname{Using}(1))\)
KCET - 2013
PHXI06:WORK ENERGY AND POWER
355696
A pendulum of length 2 \(m\) left at \(A\). When it reaches \(B\), it looses \(10 \%\) of its total energy due to air resistance. The velocity at \(B\) is
355694
Work done in time \(t\) on a body of mass \(m\) which is acceleration from rest to speed \(v\) in time \(t_{1}\) as a function of time \(t\) is given by:
1 \(\dfrac{1}{2} m v^{2}\left(\dfrac{t}{t_{1}}\right)^{2}\)
\(a=v / t_{1}\), Displacement after time \(t\): \(S=\dfrac{1}{2} a t^{2}=\dfrac{1}{2} \dfrac{v}{t_{1}} t^{2}\) \(W=F S=m a S=m \dfrac{v}{t_{1}} \dfrac{1}{2} \dfrac{v}{t_{1}} t^{2}=\dfrac{1}{2} m v^{2}\left(\dfrac{t}{t_{1}}\right)^{2}\)
PHXI06:WORK ENERGY AND POWER
355695
A truck accelerates from speed \(v\) to 2 \(v\). Work done in during this is
1 Three times as the work done in accelerating it from rest to \(v\)
2 Same as the work done in accelerating it from rest to \(v\)
3 Four times as the work done in accelerating it from rest to \(v\)
4 Lless than the work done in accelerating it from rest to \(v\)
Explanation:
Let \(m\) be mass of the truck. According to work-energy theorem \(W = {K_f} - {K_i}\) Work done in accelerating the truck from rest to speed \(v\) is \(\begin{aligned}& W_{1}=\dfrac{1}{2} m v^{2}-0 \quad(\because u=0) \\& =\dfrac{1}{2} m v^{2}\end{aligned}\) Work done in accelerating the truck from speed \(v\) to 2\(v\) is \(W_{2}=\dfrac{1}{2} m(2 v)^{2}-\dfrac{1}{2} m(v)^{2}\) \(=\dfrac{3}{2} m v^{2}=3 \dfrac{1}{2} m v^{2}=3 W_{1} \quad(\operatorname{Using}(1))\)
KCET - 2013
PHXI06:WORK ENERGY AND POWER
355696
A pendulum of length 2 \(m\) left at \(A\). When it reaches \(B\), it looses \(10 \%\) of its total energy due to air resistance. The velocity at \(B\) is
