363536
A force of \(\left( {6\hat i + 8\hat j} \right)N\) acted on a body of mass 10\(kg\). The displacement after 10sec, if it starts from rest, will be-
1 70 \(m\) along \({\tan ^{ - 1}}3/4\) with \(x\) axis
2 50 \(m\) along \({\tan ^{ - 1}}4/3\) with \(x\) axis
3 10 \(m\) along \({\tan ^{ - 1}}4/3\) with \(x\) axis
4 None
Explanation:
Acceleration \(\frac{{\overrightarrow F }}{m} = \frac{{6\hat i + 8\hat j}}{{10}}\) in the direction of the force and displacement \(\vec s = \vec ut + \frac{1}{2}\vec a{t^2} = 0 + \frac{1}{2}\left( {\frac{{6\hat i + 8\hat j}}{{10}}} \right)100 = 30\hat i + 40\hat j\) So the displace ment is 50 \(m\) along \({\tan ^{ - 1}}\frac{4}{3}\) with \(x\)-axis
PHXI05:LAWS OF MOTION
363537
A body, under the action of a force \(F = 6\hat i - 8\hat j + 10\hat k,\) acquires an acceleration of \(1\,m{s^{ - 2}}.\) The mass of this body must be
363536
A force of \(\left( {6\hat i + 8\hat j} \right)N\) acted on a body of mass 10\(kg\). The displacement after 10sec, if it starts from rest, will be-
1 70 \(m\) along \({\tan ^{ - 1}}3/4\) with \(x\) axis
2 50 \(m\) along \({\tan ^{ - 1}}4/3\) with \(x\) axis
3 10 \(m\) along \({\tan ^{ - 1}}4/3\) with \(x\) axis
4 None
Explanation:
Acceleration \(\frac{{\overrightarrow F }}{m} = \frac{{6\hat i + 8\hat j}}{{10}}\) in the direction of the force and displacement \(\vec s = \vec ut + \frac{1}{2}\vec a{t^2} = 0 + \frac{1}{2}\left( {\frac{{6\hat i + 8\hat j}}{{10}}} \right)100 = 30\hat i + 40\hat j\) So the displace ment is 50 \(m\) along \({\tan ^{ - 1}}\frac{4}{3}\) with \(x\)-axis
PHXI05:LAWS OF MOTION
363537
A body, under the action of a force \(F = 6\hat i - 8\hat j + 10\hat k,\) acquires an acceleration of \(1\,m{s^{ - 2}}.\) The mass of this body must be
363536
A force of \(\left( {6\hat i + 8\hat j} \right)N\) acted on a body of mass 10\(kg\). The displacement after 10sec, if it starts from rest, will be-
1 70 \(m\) along \({\tan ^{ - 1}}3/4\) with \(x\) axis
2 50 \(m\) along \({\tan ^{ - 1}}4/3\) with \(x\) axis
3 10 \(m\) along \({\tan ^{ - 1}}4/3\) with \(x\) axis
4 None
Explanation:
Acceleration \(\frac{{\overrightarrow F }}{m} = \frac{{6\hat i + 8\hat j}}{{10}}\) in the direction of the force and displacement \(\vec s = \vec ut + \frac{1}{2}\vec a{t^2} = 0 + \frac{1}{2}\left( {\frac{{6\hat i + 8\hat j}}{{10}}} \right)100 = 30\hat i + 40\hat j\) So the displace ment is 50 \(m\) along \({\tan ^{ - 1}}\frac{4}{3}\) with \(x\)-axis
PHXI05:LAWS OF MOTION
363537
A body, under the action of a force \(F = 6\hat i - 8\hat j + 10\hat k,\) acquires an acceleration of \(1\,m{s^{ - 2}}.\) The mass of this body must be
NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXI05:LAWS OF MOTION
363536
A force of \(\left( {6\hat i + 8\hat j} \right)N\) acted on a body of mass 10\(kg\). The displacement after 10sec, if it starts from rest, will be-
1 70 \(m\) along \({\tan ^{ - 1}}3/4\) with \(x\) axis
2 50 \(m\) along \({\tan ^{ - 1}}4/3\) with \(x\) axis
3 10 \(m\) along \({\tan ^{ - 1}}4/3\) with \(x\) axis
4 None
Explanation:
Acceleration \(\frac{{\overrightarrow F }}{m} = \frac{{6\hat i + 8\hat j}}{{10}}\) in the direction of the force and displacement \(\vec s = \vec ut + \frac{1}{2}\vec a{t^2} = 0 + \frac{1}{2}\left( {\frac{{6\hat i + 8\hat j}}{{10}}} \right)100 = 30\hat i + 40\hat j\) So the displace ment is 50 \(m\) along \({\tan ^{ - 1}}\frac{4}{3}\) with \(x\)-axis
PHXI05:LAWS OF MOTION
363537
A body, under the action of a force \(F = 6\hat i - 8\hat j + 10\hat k,\) acquires an acceleration of \(1\,m{s^{ - 2}}.\) The mass of this body must be
