NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
PHXI06:WORK ENERGY AND POWER
355611
A mass of 0.5 \(kg\) moving with a speed of 15 \(m/s\) on a horizontal smooth surface, collides with a nearly weightless spring of force constant \(k = 50\,N/m\). The maximum compression of the spring would be:
1 0.15 \(m\)
2 0.5 \(m\)
3 1.5 \(m\)
4 0.12 \(m\)
Explanation:
\(\dfrac{1}{2} m v^{2}=\dfrac{1}{2} k x^{2}\) \(\begin{aligned}& \Rightarrow m v^{2}=k x^{2} \Rightarrow 0.5 \times(1.5)^{2}=50 \times x^{2} \\& \therefore x=0.15 m\end{aligned}\)
PHXI06:WORK ENERGY AND POWER
355612
The spring extends by \(x\) on loading, then energy stored by the spring is : (if \(T\) is the tension in spring and \(k\) is spring constant)
1 \(\dfrac{2 T^{2}}{k}\)
2 \(\dfrac{T^{2}}{2 k^{2}}\)
3 \(\dfrac{T^{2}}{2 k}\)
4 \(\dfrac{2 k}{T^{2}}\)
Explanation:
\(U=\dfrac{F^{2}}{2 k}=\dfrac{T^{2}}{2 k}\).
PHXI06:WORK ENERGY AND POWER
355613
An elastic string of unstretched length \(L\) and force constant \(k\) is stretched by another small \(x\). It is further stretched by another small length \(y\). The work done in the second stretching is
1 \(\dfrac{1}{2} k y(2 x+y)\)
2 \(\dfrac{1}{2} k y^{2}\)
3 \(\dfrac{1}{2} k\left(x^{2}+y^{2}\right)\)
4 \(\dfrac{1}{2} k(x+y)^{2}\)
Explanation:
\(W=\dfrac{1}{2} k x^{2}-\dfrac{1}{2} k(x+y)^{2}\) \(W=\dfrac{1}{2} k x^{2}-\dfrac{1}{2} k\left(x^{2}+y^{2}+2 x y\right)\) \(=\dfrac{1}{2} k y(-2 x-y)\) The work done against elastic force is \(W_{e x t}=-W=\dfrac{k y}{2}(2 x+y)\)
PHXI06:WORK ENERGY AND POWER
355614
A string of length \(L\) and force constant \(K\) is stretched to obtain extension \(l\). It is further stretched to obtain extension \(l_{1}\). The work done in second stretching is
1 \(\dfrac{1}{2} K l_{1}^{2}\)
2 \(\dfrac{1}{2} K\left(l_{1}^{2}-l^{2}\right)\)
3 \(\dfrac{1}{2} K\left(l_{1}^{2}+l^{2}\right)\)
4 \(\dfrac{1}{2} K l_{1}\left(2 l+l_{1}\right)\)
Explanation:
Work done in stretching a string to obtain an extension \(l\) is \(W_{1}=\dfrac{1}{2} K l\) Similarly, work done in stretching a string to obtain extension \(l_{1}\) is \(W_{2}=\dfrac{1}{2} K l_{1}^{2}\) \(\therefore\) Work done in second case \(=W_{2}-W_{1}=\dfrac{1}{2} K\left(l_{1}^{2}-l^{2}\right)\)
355611
A mass of 0.5 \(kg\) moving with a speed of 15 \(m/s\) on a horizontal smooth surface, collides with a nearly weightless spring of force constant \(k = 50\,N/m\). The maximum compression of the spring would be:
1 0.15 \(m\)
2 0.5 \(m\)
3 1.5 \(m\)
4 0.12 \(m\)
Explanation:
\(\dfrac{1}{2} m v^{2}=\dfrac{1}{2} k x^{2}\) \(\begin{aligned}& \Rightarrow m v^{2}=k x^{2} \Rightarrow 0.5 \times(1.5)^{2}=50 \times x^{2} \\& \therefore x=0.15 m\end{aligned}\)
PHXI06:WORK ENERGY AND POWER
355612
The spring extends by \(x\) on loading, then energy stored by the spring is : (if \(T\) is the tension in spring and \(k\) is spring constant)
1 \(\dfrac{2 T^{2}}{k}\)
2 \(\dfrac{T^{2}}{2 k^{2}}\)
3 \(\dfrac{T^{2}}{2 k}\)
4 \(\dfrac{2 k}{T^{2}}\)
Explanation:
\(U=\dfrac{F^{2}}{2 k}=\dfrac{T^{2}}{2 k}\).
PHXI06:WORK ENERGY AND POWER
355613
An elastic string of unstretched length \(L\) and force constant \(k\) is stretched by another small \(x\). It is further stretched by another small length \(y\). The work done in the second stretching is
1 \(\dfrac{1}{2} k y(2 x+y)\)
2 \(\dfrac{1}{2} k y^{2}\)
3 \(\dfrac{1}{2} k\left(x^{2}+y^{2}\right)\)
4 \(\dfrac{1}{2} k(x+y)^{2}\)
Explanation:
\(W=\dfrac{1}{2} k x^{2}-\dfrac{1}{2} k(x+y)^{2}\) \(W=\dfrac{1}{2} k x^{2}-\dfrac{1}{2} k\left(x^{2}+y^{2}+2 x y\right)\) \(=\dfrac{1}{2} k y(-2 x-y)\) The work done against elastic force is \(W_{e x t}=-W=\dfrac{k y}{2}(2 x+y)\)
PHXI06:WORK ENERGY AND POWER
355614
A string of length \(L\) and force constant \(K\) is stretched to obtain extension \(l\). It is further stretched to obtain extension \(l_{1}\). The work done in second stretching is
1 \(\dfrac{1}{2} K l_{1}^{2}\)
2 \(\dfrac{1}{2} K\left(l_{1}^{2}-l^{2}\right)\)
3 \(\dfrac{1}{2} K\left(l_{1}^{2}+l^{2}\right)\)
4 \(\dfrac{1}{2} K l_{1}\left(2 l+l_{1}\right)\)
Explanation:
Work done in stretching a string to obtain an extension \(l\) is \(W_{1}=\dfrac{1}{2} K l\) Similarly, work done in stretching a string to obtain extension \(l_{1}\) is \(W_{2}=\dfrac{1}{2} K l_{1}^{2}\) \(\therefore\) Work done in second case \(=W_{2}-W_{1}=\dfrac{1}{2} K\left(l_{1}^{2}-l^{2}\right)\)
355611
A mass of 0.5 \(kg\) moving with a speed of 15 \(m/s\) on a horizontal smooth surface, collides with a nearly weightless spring of force constant \(k = 50\,N/m\). The maximum compression of the spring would be:
1 0.15 \(m\)
2 0.5 \(m\)
3 1.5 \(m\)
4 0.12 \(m\)
Explanation:
\(\dfrac{1}{2} m v^{2}=\dfrac{1}{2} k x^{2}\) \(\begin{aligned}& \Rightarrow m v^{2}=k x^{2} \Rightarrow 0.5 \times(1.5)^{2}=50 \times x^{2} \\& \therefore x=0.15 m\end{aligned}\)
PHXI06:WORK ENERGY AND POWER
355612
The spring extends by \(x\) on loading, then energy stored by the spring is : (if \(T\) is the tension in spring and \(k\) is spring constant)
1 \(\dfrac{2 T^{2}}{k}\)
2 \(\dfrac{T^{2}}{2 k^{2}}\)
3 \(\dfrac{T^{2}}{2 k}\)
4 \(\dfrac{2 k}{T^{2}}\)
Explanation:
\(U=\dfrac{F^{2}}{2 k}=\dfrac{T^{2}}{2 k}\).
PHXI06:WORK ENERGY AND POWER
355613
An elastic string of unstretched length \(L\) and force constant \(k\) is stretched by another small \(x\). It is further stretched by another small length \(y\). The work done in the second stretching is
1 \(\dfrac{1}{2} k y(2 x+y)\)
2 \(\dfrac{1}{2} k y^{2}\)
3 \(\dfrac{1}{2} k\left(x^{2}+y^{2}\right)\)
4 \(\dfrac{1}{2} k(x+y)^{2}\)
Explanation:
\(W=\dfrac{1}{2} k x^{2}-\dfrac{1}{2} k(x+y)^{2}\) \(W=\dfrac{1}{2} k x^{2}-\dfrac{1}{2} k\left(x^{2}+y^{2}+2 x y\right)\) \(=\dfrac{1}{2} k y(-2 x-y)\) The work done against elastic force is \(W_{e x t}=-W=\dfrac{k y}{2}(2 x+y)\)
PHXI06:WORK ENERGY AND POWER
355614
A string of length \(L\) and force constant \(K\) is stretched to obtain extension \(l\). It is further stretched to obtain extension \(l_{1}\). The work done in second stretching is
1 \(\dfrac{1}{2} K l_{1}^{2}\)
2 \(\dfrac{1}{2} K\left(l_{1}^{2}-l^{2}\right)\)
3 \(\dfrac{1}{2} K\left(l_{1}^{2}+l^{2}\right)\)
4 \(\dfrac{1}{2} K l_{1}\left(2 l+l_{1}\right)\)
Explanation:
Work done in stretching a string to obtain an extension \(l\) is \(W_{1}=\dfrac{1}{2} K l\) Similarly, work done in stretching a string to obtain extension \(l_{1}\) is \(W_{2}=\dfrac{1}{2} K l_{1}^{2}\) \(\therefore\) Work done in second case \(=W_{2}-W_{1}=\dfrac{1}{2} K\left(l_{1}^{2}-l^{2}\right)\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
PHXI06:WORK ENERGY AND POWER
355611
A mass of 0.5 \(kg\) moving with a speed of 15 \(m/s\) on a horizontal smooth surface, collides with a nearly weightless spring of force constant \(k = 50\,N/m\). The maximum compression of the spring would be:
1 0.15 \(m\)
2 0.5 \(m\)
3 1.5 \(m\)
4 0.12 \(m\)
Explanation:
\(\dfrac{1}{2} m v^{2}=\dfrac{1}{2} k x^{2}\) \(\begin{aligned}& \Rightarrow m v^{2}=k x^{2} \Rightarrow 0.5 \times(1.5)^{2}=50 \times x^{2} \\& \therefore x=0.15 m\end{aligned}\)
PHXI06:WORK ENERGY AND POWER
355612
The spring extends by \(x\) on loading, then energy stored by the spring is : (if \(T\) is the tension in spring and \(k\) is spring constant)
1 \(\dfrac{2 T^{2}}{k}\)
2 \(\dfrac{T^{2}}{2 k^{2}}\)
3 \(\dfrac{T^{2}}{2 k}\)
4 \(\dfrac{2 k}{T^{2}}\)
Explanation:
\(U=\dfrac{F^{2}}{2 k}=\dfrac{T^{2}}{2 k}\).
PHXI06:WORK ENERGY AND POWER
355613
An elastic string of unstretched length \(L\) and force constant \(k\) is stretched by another small \(x\). It is further stretched by another small length \(y\). The work done in the second stretching is
1 \(\dfrac{1}{2} k y(2 x+y)\)
2 \(\dfrac{1}{2} k y^{2}\)
3 \(\dfrac{1}{2} k\left(x^{2}+y^{2}\right)\)
4 \(\dfrac{1}{2} k(x+y)^{2}\)
Explanation:
\(W=\dfrac{1}{2} k x^{2}-\dfrac{1}{2} k(x+y)^{2}\) \(W=\dfrac{1}{2} k x^{2}-\dfrac{1}{2} k\left(x^{2}+y^{2}+2 x y\right)\) \(=\dfrac{1}{2} k y(-2 x-y)\) The work done against elastic force is \(W_{e x t}=-W=\dfrac{k y}{2}(2 x+y)\)
PHXI06:WORK ENERGY AND POWER
355614
A string of length \(L\) and force constant \(K\) is stretched to obtain extension \(l\). It is further stretched to obtain extension \(l_{1}\). The work done in second stretching is
1 \(\dfrac{1}{2} K l_{1}^{2}\)
2 \(\dfrac{1}{2} K\left(l_{1}^{2}-l^{2}\right)\)
3 \(\dfrac{1}{2} K\left(l_{1}^{2}+l^{2}\right)\)
4 \(\dfrac{1}{2} K l_{1}\left(2 l+l_{1}\right)\)
Explanation:
Work done in stretching a string to obtain an extension \(l\) is \(W_{1}=\dfrac{1}{2} K l\) Similarly, work done in stretching a string to obtain extension \(l_{1}\) is \(W_{2}=\dfrac{1}{2} K l_{1}^{2}\) \(\therefore\) Work done in second case \(=W_{2}-W_{1}=\dfrac{1}{2} K\left(l_{1}^{2}-l^{2}\right)\)
