371740
A ship of mass \(2 \times 10^{7} \mathrm{~kg}\) initially at rest is pulled by a force of \(5 \times 10^{5} \mathrm{~N}\) through a distance of \(2 \mathrm{~m}\). Assuming that the resistance due to water is negligible, the speed of the ship is
1 \(2 \mathrm{~ms}^{-1}\)
2 \(0.01 \mathrm{~ms}^{-1}\)
3 \(0.1 \mathrm{~ms}^{-1}\)
4 \(1 \mathrm{~ms}^{-1}\)
5 \(5 \mathrm{~ms}^{-1}\)
Explanation:
C Given, \(\mathrm{m}=2 \times 10^{7} \mathrm{~kg}\) \(\mathrm{~F}=5 \times 10^{5}\) Acceleration of the ship, applying Newton's law i.e., \(\mathrm{F}=\) ma \(\mathrm{a}=\frac{\mathrm{F}}{\mathrm{m}}=\frac{5 \times 10^{5}}{2 \times 10^{7}}=\frac{5}{2} \times 10^{-2} \mathrm{~m} / \mathrm{s}^{2}\) Final Speed of the ship, Using \(3^{\text {rd }}\) equation of motion \(v^{2}-u^{2}=2 a s\) \(v^{2}-0^{2}=2 \times \frac{5}{2} \times 10^{-2} \times 2\) \(v=0.1 \mathrm{~m} / \mathrm{s}\)
Kerala CEE 2020
LAWS OF MOTION (ADDITIONAL)
371741
A force of \((2 \hat{i}+3 \hat{j}) N\) acts on a body of mass 1 \(\mathrm{kg}\) which is at rest initially. The acceleration of the body is
B According to Newton's laws of motion, \(\mathrm{F}=\mathrm{ma} \Rightarrow \mathrm{a}=\frac{\mathrm{F}}{\mathrm{m}}\) Here, \(\quad F=(2 \hat{i}+3 \hat{j}) N\) \(\mathrm{m}=1 \mathrm{~kg}\) then, \(\quad a=\frac{(2 \hat{i}+3 \hat{j})}{1}\) \(\mathrm{a}=(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}) \mathrm{ms}^{-2}\)
Kerala CEE 2020
LAWS OF MOTION (ADDITIONAL)
371742
A constant horizontal force \(\vec{F}\) of \(30 \mathbf{N}\) is applied to block \(A\) of mass \(10 \mathrm{~kg}\) which pushes against block \(B\) of mass \(5 \mathrm{~kg}\). What is the net force on the block \(A\) ? The block are placed on a frictionless table.
1 \(20 \overline{\mathrm{N}}\)
2 \(15 \mathrm{~N}\)
3 \(10 \mathrm{~N}\)
4 \(5 \mathrm{~N}\)
Explanation:
A Since, the surface is frictionless, both the blocks will moves with a common acceleration. Acceleration, \(a=\frac{F_{\text {net }}}{m_{\text {total }}} \quad\left[F_{\text {net }}=30 \mathrm{~N}, m_{\text {total }}=15 \mathrm{~kg}\right]\) \(a =\frac{30}{(10+5)}\) \(=\frac{30}{15}\) \(=2 \mathrm{~m} / \mathrm{s}^{2}\) For, block A \(10 \mathrm{~kg} \rightarrow \mathrm{a}=2 \mathrm{~m} / \mathrm{s}^{2}\) Then, Net force on the block 'A' \(\left(\mathrm{F}_{\mathrm{net}}\right)_{\mathrm{A}} =\mathrm{m}_{\mathrm{A}} \cdot \mathrm{a}\) \(=10 \times 2\) \(=20 \mathrm{~N}\) Also for block 'B' \(\left(\mathrm{F}_{\text {net }}\right)_{\mathrm{B}} =\mathrm{m}_{\mathrm{B}} \cdot \mathrm{a}\) \(=5 \times 2\) \(=10 \mathrm{~N}\)
TS EAMCET 29.09.2020
LAWS OF MOTION (ADDITIONAL)
371743
The minimum and maximum heights attained by a child on a swing from the ground are 0.75 \(\mathrm{m}\) and \(\mathbf{2 m}\) respectively. Find his/her maximum speed:
1 \(10 \mathrm{~m} \cdot \mathrm{s}^{-1}\)
2 \(5 \mathrm{~m} \cdot \mathrm{s}^{-1}\)
3 \(8 \mathrm{~m} \cdot \mathrm{s}^{-1}\)
4 \(15 \mathrm{~m} \cdot \mathrm{s}^{-1}\)
Explanation:
B From energy conservation, \(\text { Ground }\) From energy conservation, \(\frac{1}{2} \mathrm{mv}_{\max }^{2}=\mathrm{mg}\left(\mathrm{h}_{2}-\mathrm{h}_{1}\right)\) \(\mathrm{v}_{\max }=\sqrt{2 \mathrm{~g}\left(\mathrm{~h}_{2}-\mathrm{h}_{1}\right)}\) \(\quad=\sqrt{2 \times 10(2-0.75)}=5 \mathrm{~m} / \mathrm{s}\)
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LAWS OF MOTION (ADDITIONAL)
371740
A ship of mass \(2 \times 10^{7} \mathrm{~kg}\) initially at rest is pulled by a force of \(5 \times 10^{5} \mathrm{~N}\) through a distance of \(2 \mathrm{~m}\). Assuming that the resistance due to water is negligible, the speed of the ship is
1 \(2 \mathrm{~ms}^{-1}\)
2 \(0.01 \mathrm{~ms}^{-1}\)
3 \(0.1 \mathrm{~ms}^{-1}\)
4 \(1 \mathrm{~ms}^{-1}\)
5 \(5 \mathrm{~ms}^{-1}\)
Explanation:
C Given, \(\mathrm{m}=2 \times 10^{7} \mathrm{~kg}\) \(\mathrm{~F}=5 \times 10^{5}\) Acceleration of the ship, applying Newton's law i.e., \(\mathrm{F}=\) ma \(\mathrm{a}=\frac{\mathrm{F}}{\mathrm{m}}=\frac{5 \times 10^{5}}{2 \times 10^{7}}=\frac{5}{2} \times 10^{-2} \mathrm{~m} / \mathrm{s}^{2}\) Final Speed of the ship, Using \(3^{\text {rd }}\) equation of motion \(v^{2}-u^{2}=2 a s\) \(v^{2}-0^{2}=2 \times \frac{5}{2} \times 10^{-2} \times 2\) \(v=0.1 \mathrm{~m} / \mathrm{s}\)
Kerala CEE 2020
LAWS OF MOTION (ADDITIONAL)
371741
A force of \((2 \hat{i}+3 \hat{j}) N\) acts on a body of mass 1 \(\mathrm{kg}\) which is at rest initially. The acceleration of the body is
B According to Newton's laws of motion, \(\mathrm{F}=\mathrm{ma} \Rightarrow \mathrm{a}=\frac{\mathrm{F}}{\mathrm{m}}\) Here, \(\quad F=(2 \hat{i}+3 \hat{j}) N\) \(\mathrm{m}=1 \mathrm{~kg}\) then, \(\quad a=\frac{(2 \hat{i}+3 \hat{j})}{1}\) \(\mathrm{a}=(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}) \mathrm{ms}^{-2}\)
Kerala CEE 2020
LAWS OF MOTION (ADDITIONAL)
371742
A constant horizontal force \(\vec{F}\) of \(30 \mathbf{N}\) is applied to block \(A\) of mass \(10 \mathrm{~kg}\) which pushes against block \(B\) of mass \(5 \mathrm{~kg}\). What is the net force on the block \(A\) ? The block are placed on a frictionless table.
1 \(20 \overline{\mathrm{N}}\)
2 \(15 \mathrm{~N}\)
3 \(10 \mathrm{~N}\)
4 \(5 \mathrm{~N}\)
Explanation:
A Since, the surface is frictionless, both the blocks will moves with a common acceleration. Acceleration, \(a=\frac{F_{\text {net }}}{m_{\text {total }}} \quad\left[F_{\text {net }}=30 \mathrm{~N}, m_{\text {total }}=15 \mathrm{~kg}\right]\) \(a =\frac{30}{(10+5)}\) \(=\frac{30}{15}\) \(=2 \mathrm{~m} / \mathrm{s}^{2}\) For, block A \(10 \mathrm{~kg} \rightarrow \mathrm{a}=2 \mathrm{~m} / \mathrm{s}^{2}\) Then, Net force on the block 'A' \(\left(\mathrm{F}_{\mathrm{net}}\right)_{\mathrm{A}} =\mathrm{m}_{\mathrm{A}} \cdot \mathrm{a}\) \(=10 \times 2\) \(=20 \mathrm{~N}\) Also for block 'B' \(\left(\mathrm{F}_{\text {net }}\right)_{\mathrm{B}} =\mathrm{m}_{\mathrm{B}} \cdot \mathrm{a}\) \(=5 \times 2\) \(=10 \mathrm{~N}\)
TS EAMCET 29.09.2020
LAWS OF MOTION (ADDITIONAL)
371743
The minimum and maximum heights attained by a child on a swing from the ground are 0.75 \(\mathrm{m}\) and \(\mathbf{2 m}\) respectively. Find his/her maximum speed:
1 \(10 \mathrm{~m} \cdot \mathrm{s}^{-1}\)
2 \(5 \mathrm{~m} \cdot \mathrm{s}^{-1}\)
3 \(8 \mathrm{~m} \cdot \mathrm{s}^{-1}\)
4 \(15 \mathrm{~m} \cdot \mathrm{s}^{-1}\)
Explanation:
B From energy conservation, \(\text { Ground }\) From energy conservation, \(\frac{1}{2} \mathrm{mv}_{\max }^{2}=\mathrm{mg}\left(\mathrm{h}_{2}-\mathrm{h}_{1}\right)\) \(\mathrm{v}_{\max }=\sqrt{2 \mathrm{~g}\left(\mathrm{~h}_{2}-\mathrm{h}_{1}\right)}\) \(\quad=\sqrt{2 \times 10(2-0.75)}=5 \mathrm{~m} / \mathrm{s}\)
371740
A ship of mass \(2 \times 10^{7} \mathrm{~kg}\) initially at rest is pulled by a force of \(5 \times 10^{5} \mathrm{~N}\) through a distance of \(2 \mathrm{~m}\). Assuming that the resistance due to water is negligible, the speed of the ship is
1 \(2 \mathrm{~ms}^{-1}\)
2 \(0.01 \mathrm{~ms}^{-1}\)
3 \(0.1 \mathrm{~ms}^{-1}\)
4 \(1 \mathrm{~ms}^{-1}\)
5 \(5 \mathrm{~ms}^{-1}\)
Explanation:
C Given, \(\mathrm{m}=2 \times 10^{7} \mathrm{~kg}\) \(\mathrm{~F}=5 \times 10^{5}\) Acceleration of the ship, applying Newton's law i.e., \(\mathrm{F}=\) ma \(\mathrm{a}=\frac{\mathrm{F}}{\mathrm{m}}=\frac{5 \times 10^{5}}{2 \times 10^{7}}=\frac{5}{2} \times 10^{-2} \mathrm{~m} / \mathrm{s}^{2}\) Final Speed of the ship, Using \(3^{\text {rd }}\) equation of motion \(v^{2}-u^{2}=2 a s\) \(v^{2}-0^{2}=2 \times \frac{5}{2} \times 10^{-2} \times 2\) \(v=0.1 \mathrm{~m} / \mathrm{s}\)
Kerala CEE 2020
LAWS OF MOTION (ADDITIONAL)
371741
A force of \((2 \hat{i}+3 \hat{j}) N\) acts on a body of mass 1 \(\mathrm{kg}\) which is at rest initially. The acceleration of the body is
B According to Newton's laws of motion, \(\mathrm{F}=\mathrm{ma} \Rightarrow \mathrm{a}=\frac{\mathrm{F}}{\mathrm{m}}\) Here, \(\quad F=(2 \hat{i}+3 \hat{j}) N\) \(\mathrm{m}=1 \mathrm{~kg}\) then, \(\quad a=\frac{(2 \hat{i}+3 \hat{j})}{1}\) \(\mathrm{a}=(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}) \mathrm{ms}^{-2}\)
Kerala CEE 2020
LAWS OF MOTION (ADDITIONAL)
371742
A constant horizontal force \(\vec{F}\) of \(30 \mathbf{N}\) is applied to block \(A\) of mass \(10 \mathrm{~kg}\) which pushes against block \(B\) of mass \(5 \mathrm{~kg}\). What is the net force on the block \(A\) ? The block are placed on a frictionless table.
1 \(20 \overline{\mathrm{N}}\)
2 \(15 \mathrm{~N}\)
3 \(10 \mathrm{~N}\)
4 \(5 \mathrm{~N}\)
Explanation:
A Since, the surface is frictionless, both the blocks will moves with a common acceleration. Acceleration, \(a=\frac{F_{\text {net }}}{m_{\text {total }}} \quad\left[F_{\text {net }}=30 \mathrm{~N}, m_{\text {total }}=15 \mathrm{~kg}\right]\) \(a =\frac{30}{(10+5)}\) \(=\frac{30}{15}\) \(=2 \mathrm{~m} / \mathrm{s}^{2}\) For, block A \(10 \mathrm{~kg} \rightarrow \mathrm{a}=2 \mathrm{~m} / \mathrm{s}^{2}\) Then, Net force on the block 'A' \(\left(\mathrm{F}_{\mathrm{net}}\right)_{\mathrm{A}} =\mathrm{m}_{\mathrm{A}} \cdot \mathrm{a}\) \(=10 \times 2\) \(=20 \mathrm{~N}\) Also for block 'B' \(\left(\mathrm{F}_{\text {net }}\right)_{\mathrm{B}} =\mathrm{m}_{\mathrm{B}} \cdot \mathrm{a}\) \(=5 \times 2\) \(=10 \mathrm{~N}\)
TS EAMCET 29.09.2020
LAWS OF MOTION (ADDITIONAL)
371743
The minimum and maximum heights attained by a child on a swing from the ground are 0.75 \(\mathrm{m}\) and \(\mathbf{2 m}\) respectively. Find his/her maximum speed:
1 \(10 \mathrm{~m} \cdot \mathrm{s}^{-1}\)
2 \(5 \mathrm{~m} \cdot \mathrm{s}^{-1}\)
3 \(8 \mathrm{~m} \cdot \mathrm{s}^{-1}\)
4 \(15 \mathrm{~m} \cdot \mathrm{s}^{-1}\)
Explanation:
B From energy conservation, \(\text { Ground }\) From energy conservation, \(\frac{1}{2} \mathrm{mv}_{\max }^{2}=\mathrm{mg}\left(\mathrm{h}_{2}-\mathrm{h}_{1}\right)\) \(\mathrm{v}_{\max }=\sqrt{2 \mathrm{~g}\left(\mathrm{~h}_{2}-\mathrm{h}_{1}\right)}\) \(\quad=\sqrt{2 \times 10(2-0.75)}=5 \mathrm{~m} / \mathrm{s}\)
371740
A ship of mass \(2 \times 10^{7} \mathrm{~kg}\) initially at rest is pulled by a force of \(5 \times 10^{5} \mathrm{~N}\) through a distance of \(2 \mathrm{~m}\). Assuming that the resistance due to water is negligible, the speed of the ship is
1 \(2 \mathrm{~ms}^{-1}\)
2 \(0.01 \mathrm{~ms}^{-1}\)
3 \(0.1 \mathrm{~ms}^{-1}\)
4 \(1 \mathrm{~ms}^{-1}\)
5 \(5 \mathrm{~ms}^{-1}\)
Explanation:
C Given, \(\mathrm{m}=2 \times 10^{7} \mathrm{~kg}\) \(\mathrm{~F}=5 \times 10^{5}\) Acceleration of the ship, applying Newton's law i.e., \(\mathrm{F}=\) ma \(\mathrm{a}=\frac{\mathrm{F}}{\mathrm{m}}=\frac{5 \times 10^{5}}{2 \times 10^{7}}=\frac{5}{2} \times 10^{-2} \mathrm{~m} / \mathrm{s}^{2}\) Final Speed of the ship, Using \(3^{\text {rd }}\) equation of motion \(v^{2}-u^{2}=2 a s\) \(v^{2}-0^{2}=2 \times \frac{5}{2} \times 10^{-2} \times 2\) \(v=0.1 \mathrm{~m} / \mathrm{s}\)
Kerala CEE 2020
LAWS OF MOTION (ADDITIONAL)
371741
A force of \((2 \hat{i}+3 \hat{j}) N\) acts on a body of mass 1 \(\mathrm{kg}\) which is at rest initially. The acceleration of the body is
B According to Newton's laws of motion, \(\mathrm{F}=\mathrm{ma} \Rightarrow \mathrm{a}=\frac{\mathrm{F}}{\mathrm{m}}\) Here, \(\quad F=(2 \hat{i}+3 \hat{j}) N\) \(\mathrm{m}=1 \mathrm{~kg}\) then, \(\quad a=\frac{(2 \hat{i}+3 \hat{j})}{1}\) \(\mathrm{a}=(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}) \mathrm{ms}^{-2}\)
Kerala CEE 2020
LAWS OF MOTION (ADDITIONAL)
371742
A constant horizontal force \(\vec{F}\) of \(30 \mathbf{N}\) is applied to block \(A\) of mass \(10 \mathrm{~kg}\) which pushes against block \(B\) of mass \(5 \mathrm{~kg}\). What is the net force on the block \(A\) ? The block are placed on a frictionless table.
1 \(20 \overline{\mathrm{N}}\)
2 \(15 \mathrm{~N}\)
3 \(10 \mathrm{~N}\)
4 \(5 \mathrm{~N}\)
Explanation:
A Since, the surface is frictionless, both the blocks will moves with a common acceleration. Acceleration, \(a=\frac{F_{\text {net }}}{m_{\text {total }}} \quad\left[F_{\text {net }}=30 \mathrm{~N}, m_{\text {total }}=15 \mathrm{~kg}\right]\) \(a =\frac{30}{(10+5)}\) \(=\frac{30}{15}\) \(=2 \mathrm{~m} / \mathrm{s}^{2}\) For, block A \(10 \mathrm{~kg} \rightarrow \mathrm{a}=2 \mathrm{~m} / \mathrm{s}^{2}\) Then, Net force on the block 'A' \(\left(\mathrm{F}_{\mathrm{net}}\right)_{\mathrm{A}} =\mathrm{m}_{\mathrm{A}} \cdot \mathrm{a}\) \(=10 \times 2\) \(=20 \mathrm{~N}\) Also for block 'B' \(\left(\mathrm{F}_{\text {net }}\right)_{\mathrm{B}} =\mathrm{m}_{\mathrm{B}} \cdot \mathrm{a}\) \(=5 \times 2\) \(=10 \mathrm{~N}\)
TS EAMCET 29.09.2020
LAWS OF MOTION (ADDITIONAL)
371743
The minimum and maximum heights attained by a child on a swing from the ground are 0.75 \(\mathrm{m}\) and \(\mathbf{2 m}\) respectively. Find his/her maximum speed:
1 \(10 \mathrm{~m} \cdot \mathrm{s}^{-1}\)
2 \(5 \mathrm{~m} \cdot \mathrm{s}^{-1}\)
3 \(8 \mathrm{~m} \cdot \mathrm{s}^{-1}\)
4 \(15 \mathrm{~m} \cdot \mathrm{s}^{-1}\)
Explanation:
B From energy conservation, \(\text { Ground }\) From energy conservation, \(\frac{1}{2} \mathrm{mv}_{\max }^{2}=\mathrm{mg}\left(\mathrm{h}_{2}-\mathrm{h}_{1}\right)\) \(\mathrm{v}_{\max }=\sqrt{2 \mathrm{~g}\left(\mathrm{~h}_{2}-\mathrm{h}_{1}\right)}\) \(\quad=\sqrt{2 \times 10(2-0.75)}=5 \mathrm{~m} / \mathrm{s}\)
