363399
Calculate the maximum acceleration of a moving car so that a body lying on the floor of the car remains stationary. The coefficient of static friction between the body and the floor is 0.15 \(\left( {g = 10\;m{s^{ - 2}}} \right)\)
1 \(150\;m{s^{ - 2}}\)
2 \(1.5\;m{s^{ - 2}}\)
3 \(50\;m{s^{ - 2}}\)
4 \(1.2\;m{s^{ - 2}}\)
Explanation:
\(f_{L}=m a_{\max }\) \(\Rightarrow \mu m g=m a_{\max }\) \( \Rightarrow {a_{\max }} = \mu g = 0.15(10) = 1.5\;m/{s^2}\). Correct option is (2).
NEET - 2023
PHXI05:LAWS OF MOTION
363400
In the figure shown, what is the minimum force (in \(N\) ) with which string should be pulled to make the block move along horizontal surface? Neglect masses of pulley and string and friction between pulley and string ( \(g=9.8 {~m} / {s}^{2}\) )
363401
A block of weight 5 \(N\) is pushed aganist a vertical wall by a force 12 \(N\). The coefficient of friction between the wall of block is 0.6. The magnitude of the force exerted by the wall on the block is
1 \({\rm{12}}\,N\)
2 \({\rm{5}}\,N\)
3 \(7.2\,N\)
4 \(13\,N\)
Explanation:
\(N = \) applied \( = 12\,N\) \(\therefore \,\,\,{f_{\max }} = \mu N = 7.2N\) Since weight \(w < {f_{\max }}\) Force of friction \(f = 5N\) \(\therefore \) Net contact force \( = \sqrt {{N^2} + {f^2}} \) \( = \sqrt {{{12}^2} + {5^2}} = 13N\)
PHXI05:LAWS OF MOTION
363402
Assertion : Pulling a lawn roller is easier than pushing it. Reason : Pushing increases the apparent weight and hence the force of friction.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
\(P = \) Pull, \(Q = \) Push. \(\theta\) made by \(P\) and \(Q\) with horizontal is same. \(M=\) Mass of lawn roller. \(f=\) Frictional force (limiting) \(\mu_{s}=\) Coefficient of friction. \(N=\) Normal reaction. We know \(f=\mu_{s} N\). Free body diagram for case of 'Pull' is as below. 'Pull' decreases value of \(M g\) by subtracting (Psin \(\theta\) ) \(f_{\text {Pull }}=\mu_{s}(M g-P \sin \theta)\) Free body diagram for case of 'Push' is as below 'Push' increases value of \(Mg\) by appending \((Q \sin \theta)\). \(f_{\text {Push }}=\mu_{s}(M g+Q \sin \theta)\) \(\Rightarrow f_{\text {push }}>f_{\text {pull }}\) (for all values of \(\mu_{s}, M, P, Q, \theta\) ) \(\Rightarrow\) It is easier to pull than pushing. The reason correctly explains this, stating that pushing increases the apparent weight and hence the force of friction. This makes it more difficult to move the roller by 'Push'. So correct option is (1)
363399
Calculate the maximum acceleration of a moving car so that a body lying on the floor of the car remains stationary. The coefficient of static friction between the body and the floor is 0.15 \(\left( {g = 10\;m{s^{ - 2}}} \right)\)
1 \(150\;m{s^{ - 2}}\)
2 \(1.5\;m{s^{ - 2}}\)
3 \(50\;m{s^{ - 2}}\)
4 \(1.2\;m{s^{ - 2}}\)
Explanation:
\(f_{L}=m a_{\max }\) \(\Rightarrow \mu m g=m a_{\max }\) \( \Rightarrow {a_{\max }} = \mu g = 0.15(10) = 1.5\;m/{s^2}\). Correct option is (2).
NEET - 2023
PHXI05:LAWS OF MOTION
363400
In the figure shown, what is the minimum force (in \(N\) ) with which string should be pulled to make the block move along horizontal surface? Neglect masses of pulley and string and friction between pulley and string ( \(g=9.8 {~m} / {s}^{2}\) )
363401
A block of weight 5 \(N\) is pushed aganist a vertical wall by a force 12 \(N\). The coefficient of friction between the wall of block is 0.6. The magnitude of the force exerted by the wall on the block is
1 \({\rm{12}}\,N\)
2 \({\rm{5}}\,N\)
3 \(7.2\,N\)
4 \(13\,N\)
Explanation:
\(N = \) applied \( = 12\,N\) \(\therefore \,\,\,{f_{\max }} = \mu N = 7.2N\) Since weight \(w < {f_{\max }}\) Force of friction \(f = 5N\) \(\therefore \) Net contact force \( = \sqrt {{N^2} + {f^2}} \) \( = \sqrt {{{12}^2} + {5^2}} = 13N\)
PHXI05:LAWS OF MOTION
363402
Assertion : Pulling a lawn roller is easier than pushing it. Reason : Pushing increases the apparent weight and hence the force of friction.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
\(P = \) Pull, \(Q = \) Push. \(\theta\) made by \(P\) and \(Q\) with horizontal is same. \(M=\) Mass of lawn roller. \(f=\) Frictional force (limiting) \(\mu_{s}=\) Coefficient of friction. \(N=\) Normal reaction. We know \(f=\mu_{s} N\). Free body diagram for case of 'Pull' is as below. 'Pull' decreases value of \(M g\) by subtracting (Psin \(\theta\) ) \(f_{\text {Pull }}=\mu_{s}(M g-P \sin \theta)\) Free body diagram for case of 'Push' is as below 'Push' increases value of \(Mg\) by appending \((Q \sin \theta)\). \(f_{\text {Push }}=\mu_{s}(M g+Q \sin \theta)\) \(\Rightarrow f_{\text {push }}>f_{\text {pull }}\) (for all values of \(\mu_{s}, M, P, Q, \theta\) ) \(\Rightarrow\) It is easier to pull than pushing. The reason correctly explains this, stating that pushing increases the apparent weight and hence the force of friction. This makes it more difficult to move the roller by 'Push'. So correct option is (1)
363399
Calculate the maximum acceleration of a moving car so that a body lying on the floor of the car remains stationary. The coefficient of static friction between the body and the floor is 0.15 \(\left( {g = 10\;m{s^{ - 2}}} \right)\)
1 \(150\;m{s^{ - 2}}\)
2 \(1.5\;m{s^{ - 2}}\)
3 \(50\;m{s^{ - 2}}\)
4 \(1.2\;m{s^{ - 2}}\)
Explanation:
\(f_{L}=m a_{\max }\) \(\Rightarrow \mu m g=m a_{\max }\) \( \Rightarrow {a_{\max }} = \mu g = 0.15(10) = 1.5\;m/{s^2}\). Correct option is (2).
NEET - 2023
PHXI05:LAWS OF MOTION
363400
In the figure shown, what is the minimum force (in \(N\) ) with which string should be pulled to make the block move along horizontal surface? Neglect masses of pulley and string and friction between pulley and string ( \(g=9.8 {~m} / {s}^{2}\) )
363401
A block of weight 5 \(N\) is pushed aganist a vertical wall by a force 12 \(N\). The coefficient of friction between the wall of block is 0.6. The magnitude of the force exerted by the wall on the block is
1 \({\rm{12}}\,N\)
2 \({\rm{5}}\,N\)
3 \(7.2\,N\)
4 \(13\,N\)
Explanation:
\(N = \) applied \( = 12\,N\) \(\therefore \,\,\,{f_{\max }} = \mu N = 7.2N\) Since weight \(w < {f_{\max }}\) Force of friction \(f = 5N\) \(\therefore \) Net contact force \( = \sqrt {{N^2} + {f^2}} \) \( = \sqrt {{{12}^2} + {5^2}} = 13N\)
PHXI05:LAWS OF MOTION
363402
Assertion : Pulling a lawn roller is easier than pushing it. Reason : Pushing increases the apparent weight and hence the force of friction.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
\(P = \) Pull, \(Q = \) Push. \(\theta\) made by \(P\) and \(Q\) with horizontal is same. \(M=\) Mass of lawn roller. \(f=\) Frictional force (limiting) \(\mu_{s}=\) Coefficient of friction. \(N=\) Normal reaction. We know \(f=\mu_{s} N\). Free body diagram for case of 'Pull' is as below. 'Pull' decreases value of \(M g\) by subtracting (Psin \(\theta\) ) \(f_{\text {Pull }}=\mu_{s}(M g-P \sin \theta)\) Free body diagram for case of 'Push' is as below 'Push' increases value of \(Mg\) by appending \((Q \sin \theta)\). \(f_{\text {Push }}=\mu_{s}(M g+Q \sin \theta)\) \(\Rightarrow f_{\text {push }}>f_{\text {pull }}\) (for all values of \(\mu_{s}, M, P, Q, \theta\) ) \(\Rightarrow\) It is easier to pull than pushing. The reason correctly explains this, stating that pushing increases the apparent weight and hence the force of friction. This makes it more difficult to move the roller by 'Push'. So correct option is (1)
363399
Calculate the maximum acceleration of a moving car so that a body lying on the floor of the car remains stationary. The coefficient of static friction between the body and the floor is 0.15 \(\left( {g = 10\;m{s^{ - 2}}} \right)\)
1 \(150\;m{s^{ - 2}}\)
2 \(1.5\;m{s^{ - 2}}\)
3 \(50\;m{s^{ - 2}}\)
4 \(1.2\;m{s^{ - 2}}\)
Explanation:
\(f_{L}=m a_{\max }\) \(\Rightarrow \mu m g=m a_{\max }\) \( \Rightarrow {a_{\max }} = \mu g = 0.15(10) = 1.5\;m/{s^2}\). Correct option is (2).
NEET - 2023
PHXI05:LAWS OF MOTION
363400
In the figure shown, what is the minimum force (in \(N\) ) with which string should be pulled to make the block move along horizontal surface? Neglect masses of pulley and string and friction between pulley and string ( \(g=9.8 {~m} / {s}^{2}\) )
363401
A block of weight 5 \(N\) is pushed aganist a vertical wall by a force 12 \(N\). The coefficient of friction between the wall of block is 0.6. The magnitude of the force exerted by the wall on the block is
1 \({\rm{12}}\,N\)
2 \({\rm{5}}\,N\)
3 \(7.2\,N\)
4 \(13\,N\)
Explanation:
\(N = \) applied \( = 12\,N\) \(\therefore \,\,\,{f_{\max }} = \mu N = 7.2N\) Since weight \(w < {f_{\max }}\) Force of friction \(f = 5N\) \(\therefore \) Net contact force \( = \sqrt {{N^2} + {f^2}} \) \( = \sqrt {{{12}^2} + {5^2}} = 13N\)
PHXI05:LAWS OF MOTION
363402
Assertion : Pulling a lawn roller is easier than pushing it. Reason : Pushing increases the apparent weight and hence the force of friction.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
\(P = \) Pull, \(Q = \) Push. \(\theta\) made by \(P\) and \(Q\) with horizontal is same. \(M=\) Mass of lawn roller. \(f=\) Frictional force (limiting) \(\mu_{s}=\) Coefficient of friction. \(N=\) Normal reaction. We know \(f=\mu_{s} N\). Free body diagram for case of 'Pull' is as below. 'Pull' decreases value of \(M g\) by subtracting (Psin \(\theta\) ) \(f_{\text {Pull }}=\mu_{s}(M g-P \sin \theta)\) Free body diagram for case of 'Push' is as below 'Push' increases value of \(Mg\) by appending \((Q \sin \theta)\). \(f_{\text {Push }}=\mu_{s}(M g+Q \sin \theta)\) \(\Rightarrow f_{\text {push }}>f_{\text {pull }}\) (for all values of \(\mu_{s}, M, P, Q, \theta\) ) \(\Rightarrow\) It is easier to pull than pushing. The reason correctly explains this, stating that pushing increases the apparent weight and hence the force of friction. This makes it more difficult to move the roller by 'Push'. So correct option is (1)
