355778
A block of mass \(100\,kg\) slides over a distance of \(10 \,m\) on a horizontal surface. If the coefficient of friction between the surfaces is 0.4 then the work done against friction (in \(J\) ) is
1 4200
2 4000
3 4500
4 3900
Explanation:
Work done against friction \(=-\) Work done by friction \(\therefore \,\,\,\,W = {f_k}s = {\mu _k}Ns\) \( = {\mu _k}mgs\) \( = 0.4 \times 100 \times 10 \times 10 = 4000\;J\)
JEE - 2024
PHXI06:WORK ENERGY AND POWER
355779
Work done by force of friction
1 Can be zero
2 Can be positive
3 Can be negative
4 Any of the above
Explanation:
Conceptual Question
PHXI06:WORK ENERGY AND POWER
355780
A weight lifter jerks \(220\;kg\) vertically through 1.5 meters and holds still at that height for two minutes. The works done by him in lifting and in holding it still are respectively
1 \(220\;\,J,330\,\;J\)
2 \(3234\;\,\,J,0\)
3 \(2334\;\,J,10\;\,J\)
4 \(0,3234\;\,J\)
Explanation:
Work done in lifting through height \(h=m g h\). No work is done in holding the weight at height \('h'\). Hence total work done \(=m g h\) \( = 220 \times 9.8 \times 1.5\) \( = 3234\,J\) Hence answers are respectively \(3234\,J\), zero.
PHXI06:WORK ENERGY AND POWER
355781
A particle is moved from a position \({\vec{r}_{1}=(3 \hat{i}+2 \hat{j}-6 \hat{k}) m}\) to a position \({\vec{r}_{2}=(14 \hat{i}+13 \hat{j}-9 \hat{k})}\) metre under the action of a force \({\vec{F}=(4 \hat{i}+\hat{j}+3 \hat{k}) N}\). The workdone is equal to
355782
A force of magnitude \(F\) acts on the free end of the cord. The weight moves up slowly by a distance \(h\). How much work is done on the weight by the rope connecting pulley and weight?
1 \(Fh/2\)
2 \(Fh\)
3 \(2Fh\)
4 None of these
Explanation:
We are applying force \(F\) therefore tension in string is equal to \(F\). Hence force acting on weight is 2 \(F\) upward, and displacement is \(h\) upward. Hence \(W = 2Fh\) (work done on weight)
355778
A block of mass \(100\,kg\) slides over a distance of \(10 \,m\) on a horizontal surface. If the coefficient of friction between the surfaces is 0.4 then the work done against friction (in \(J\) ) is
1 4200
2 4000
3 4500
4 3900
Explanation:
Work done against friction \(=-\) Work done by friction \(\therefore \,\,\,\,W = {f_k}s = {\mu _k}Ns\) \( = {\mu _k}mgs\) \( = 0.4 \times 100 \times 10 \times 10 = 4000\;J\)
JEE - 2024
PHXI06:WORK ENERGY AND POWER
355779
Work done by force of friction
1 Can be zero
2 Can be positive
3 Can be negative
4 Any of the above
Explanation:
Conceptual Question
PHXI06:WORK ENERGY AND POWER
355780
A weight lifter jerks \(220\;kg\) vertically through 1.5 meters and holds still at that height for two minutes. The works done by him in lifting and in holding it still are respectively
1 \(220\;\,J,330\,\;J\)
2 \(3234\;\,\,J,0\)
3 \(2334\;\,J,10\;\,J\)
4 \(0,3234\;\,J\)
Explanation:
Work done in lifting through height \(h=m g h\). No work is done in holding the weight at height \('h'\). Hence total work done \(=m g h\) \( = 220 \times 9.8 \times 1.5\) \( = 3234\,J\) Hence answers are respectively \(3234\,J\), zero.
PHXI06:WORK ENERGY AND POWER
355781
A particle is moved from a position \({\vec{r}_{1}=(3 \hat{i}+2 \hat{j}-6 \hat{k}) m}\) to a position \({\vec{r}_{2}=(14 \hat{i}+13 \hat{j}-9 \hat{k})}\) metre under the action of a force \({\vec{F}=(4 \hat{i}+\hat{j}+3 \hat{k}) N}\). The workdone is equal to
355782
A force of magnitude \(F\) acts on the free end of the cord. The weight moves up slowly by a distance \(h\). How much work is done on the weight by the rope connecting pulley and weight?
1 \(Fh/2\)
2 \(Fh\)
3 \(2Fh\)
4 None of these
Explanation:
We are applying force \(F\) therefore tension in string is equal to \(F\). Hence force acting on weight is 2 \(F\) upward, and displacement is \(h\) upward. Hence \(W = 2Fh\) (work done on weight)
355778
A block of mass \(100\,kg\) slides over a distance of \(10 \,m\) on a horizontal surface. If the coefficient of friction between the surfaces is 0.4 then the work done against friction (in \(J\) ) is
1 4200
2 4000
3 4500
4 3900
Explanation:
Work done against friction \(=-\) Work done by friction \(\therefore \,\,\,\,W = {f_k}s = {\mu _k}Ns\) \( = {\mu _k}mgs\) \( = 0.4 \times 100 \times 10 \times 10 = 4000\;J\)
JEE - 2024
PHXI06:WORK ENERGY AND POWER
355779
Work done by force of friction
1 Can be zero
2 Can be positive
3 Can be negative
4 Any of the above
Explanation:
Conceptual Question
PHXI06:WORK ENERGY AND POWER
355780
A weight lifter jerks \(220\;kg\) vertically through 1.5 meters and holds still at that height for two minutes. The works done by him in lifting and in holding it still are respectively
1 \(220\;\,J,330\,\;J\)
2 \(3234\;\,\,J,0\)
3 \(2334\;\,J,10\;\,J\)
4 \(0,3234\;\,J\)
Explanation:
Work done in lifting through height \(h=m g h\). No work is done in holding the weight at height \('h'\). Hence total work done \(=m g h\) \( = 220 \times 9.8 \times 1.5\) \( = 3234\,J\) Hence answers are respectively \(3234\,J\), zero.
PHXI06:WORK ENERGY AND POWER
355781
A particle is moved from a position \({\vec{r}_{1}=(3 \hat{i}+2 \hat{j}-6 \hat{k}) m}\) to a position \({\vec{r}_{2}=(14 \hat{i}+13 \hat{j}-9 \hat{k})}\) metre under the action of a force \({\vec{F}=(4 \hat{i}+\hat{j}+3 \hat{k}) N}\). The workdone is equal to
355782
A force of magnitude \(F\) acts on the free end of the cord. The weight moves up slowly by a distance \(h\). How much work is done on the weight by the rope connecting pulley and weight?
1 \(Fh/2\)
2 \(Fh\)
3 \(2Fh\)
4 None of these
Explanation:
We are applying force \(F\) therefore tension in string is equal to \(F\). Hence force acting on weight is 2 \(F\) upward, and displacement is \(h\) upward. Hence \(W = 2Fh\) (work done on weight)
355778
A block of mass \(100\,kg\) slides over a distance of \(10 \,m\) on a horizontal surface. If the coefficient of friction between the surfaces is 0.4 then the work done against friction (in \(J\) ) is
1 4200
2 4000
3 4500
4 3900
Explanation:
Work done against friction \(=-\) Work done by friction \(\therefore \,\,\,\,W = {f_k}s = {\mu _k}Ns\) \( = {\mu _k}mgs\) \( = 0.4 \times 100 \times 10 \times 10 = 4000\;J\)
JEE - 2024
PHXI06:WORK ENERGY AND POWER
355779
Work done by force of friction
1 Can be zero
2 Can be positive
3 Can be negative
4 Any of the above
Explanation:
Conceptual Question
PHXI06:WORK ENERGY AND POWER
355780
A weight lifter jerks \(220\;kg\) vertically through 1.5 meters and holds still at that height for two minutes. The works done by him in lifting and in holding it still are respectively
1 \(220\;\,J,330\,\;J\)
2 \(3234\;\,\,J,0\)
3 \(2334\;\,J,10\;\,J\)
4 \(0,3234\;\,J\)
Explanation:
Work done in lifting through height \(h=m g h\). No work is done in holding the weight at height \('h'\). Hence total work done \(=m g h\) \( = 220 \times 9.8 \times 1.5\) \( = 3234\,J\) Hence answers are respectively \(3234\,J\), zero.
PHXI06:WORK ENERGY AND POWER
355781
A particle is moved from a position \({\vec{r}_{1}=(3 \hat{i}+2 \hat{j}-6 \hat{k}) m}\) to a position \({\vec{r}_{2}=(14 \hat{i}+13 \hat{j}-9 \hat{k})}\) metre under the action of a force \({\vec{F}=(4 \hat{i}+\hat{j}+3 \hat{k}) N}\). The workdone is equal to
355782
A force of magnitude \(F\) acts on the free end of the cord. The weight moves up slowly by a distance \(h\). How much work is done on the weight by the rope connecting pulley and weight?
1 \(Fh/2\)
2 \(Fh\)
3 \(2Fh\)
4 None of these
Explanation:
We are applying force \(F\) therefore tension in string is equal to \(F\). Hence force acting on weight is 2 \(F\) upward, and displacement is \(h\) upward. Hence \(W = 2Fh\) (work done on weight)
355778
A block of mass \(100\,kg\) slides over a distance of \(10 \,m\) on a horizontal surface. If the coefficient of friction between the surfaces is 0.4 then the work done against friction (in \(J\) ) is
1 4200
2 4000
3 4500
4 3900
Explanation:
Work done against friction \(=-\) Work done by friction \(\therefore \,\,\,\,W = {f_k}s = {\mu _k}Ns\) \( = {\mu _k}mgs\) \( = 0.4 \times 100 \times 10 \times 10 = 4000\;J\)
JEE - 2024
PHXI06:WORK ENERGY AND POWER
355779
Work done by force of friction
1 Can be zero
2 Can be positive
3 Can be negative
4 Any of the above
Explanation:
Conceptual Question
PHXI06:WORK ENERGY AND POWER
355780
A weight lifter jerks \(220\;kg\) vertically through 1.5 meters and holds still at that height for two minutes. The works done by him in lifting and in holding it still are respectively
1 \(220\;\,J,330\,\;J\)
2 \(3234\;\,\,J,0\)
3 \(2334\;\,J,10\;\,J\)
4 \(0,3234\;\,J\)
Explanation:
Work done in lifting through height \(h=m g h\). No work is done in holding the weight at height \('h'\). Hence total work done \(=m g h\) \( = 220 \times 9.8 \times 1.5\) \( = 3234\,J\) Hence answers are respectively \(3234\,J\), zero.
PHXI06:WORK ENERGY AND POWER
355781
A particle is moved from a position \({\vec{r}_{1}=(3 \hat{i}+2 \hat{j}-6 \hat{k}) m}\) to a position \({\vec{r}_{2}=(14 \hat{i}+13 \hat{j}-9 \hat{k})}\) metre under the action of a force \({\vec{F}=(4 \hat{i}+\hat{j}+3 \hat{k}) N}\). The workdone is equal to
355782
A force of magnitude \(F\) acts on the free end of the cord. The weight moves up slowly by a distance \(h\). How much work is done on the weight by the rope connecting pulley and weight?
1 \(Fh/2\)
2 \(Fh\)
3 \(2Fh\)
4 None of these
Explanation:
We are applying force \(F\) therefore tension in string is equal to \(F\). Hence force acting on weight is 2 \(F\) upward, and displacement is \(h\) upward. Hence \(W = 2Fh\) (work done on weight)
