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LAWS OF MOTION (ADDITIONAL)

371770 A man of mass $60 kg$ climbed down using an elevator. The elevator had an acceleration 4 ${ms}^{- 2}$ . If the acceleration due to gravity is $10$ ${ms}^{- 2}$ , the man's apparent weight on his way down is

1

60 N

2

240 N

3

360 N

4

840 N

5

3600 N

Explanation:

C Given, $m = 60 kg, a_{0} = 4 m / s^{2}$
With respect to the frame of reference of elevator,

$\sum F = 0$
$N + {ma}_{0} = mg$
$N = m (g - a_{0})$
$N = 60 (10 - 4) = 60 \times 6$
$N = 360 N$

Kerala CEE - 2017

LAWS OF MOTION (ADDITIONAL)

371771 A man weighing $100 kg$ slides down a light rope with an acceleration of $1.8 {ms}^{- 2}$ . If $g = 9.8 {ms}^{- 2}$ , the tension of the rope is

1

180 N

2

1160 N

3

800 N

4 weightlessness

Explanation:

C Given, mass of man $(m) = 100 kg$ acceleration $= 1.8 m / s^{2}$

We know that,
$∴ T + ma = mg$
$T = m (g - a)$
$T = 100 (9.8 - 1.8) = 800 N$
Hence, the tension of the rope is $800 N$

Manipal UGET-2017

LAWS OF MOTION (ADDITIONAL)

371772 A body of mass $6 kg$ is hanging from another body of mass $10 kg$ as shown in figure. This combination is being pulled up by a string with an acceleration of $2 {ms}^{- 2}$ . The tension $T_{1}$ is, $(g =$ $10 {m s}^{- 2}$ )

1

240 N

2

150 N

3

220 N

4

192 N

5

178 N

Explanation:

D Given,
$m_{1} = 10 kg$
$m_{2} = 6 kg$
mass of system, $M = m_{1} + m_{2}$

$T_{1} = M (g + a)$
Tension is pulling the mass $(M = 10 + 6 = 16 kg)$
$T_{1} = (10 + 6) (10 + 2)$
$T_{1} = 192 N$

Kerala CEE - 2010

LAWS OF MOTION (ADDITIONAL)

371773 An object of mass $5 kg$ is attached to the hook of a spring balance and the balance is suspended vertically from the roof of a lift. The reading on the spring balance when the lift is going up with an acceleration of $0.25 {ms}^{- 2}$ is taken $(g = 10 {ms}^{- 2})$

1

51.25 N

2

48.75 N

3

52.75 N

4

47.25 N

5

55 N

Explanation:

A Given that,
$m = 5 kg$
$a = 0.25 m / s^{2}$

Let $N$ be the reading on spring Balance-

Force will be equal to the weight-
Weight $= m g + m a$
$= m (g + a)$
Weight $= 5 (10 + 0.25)$
$∴ = 51.25 N$

Kerala CEE - 2008

LAWS OF MOTION (ADDITIONAL)

371770 A man of mass $60 kg$ climbed down using an elevator. The elevator had an acceleration 4 ${ms}^{- 2}$ . If the acceleration due to gravity is $10$ ${ms}^{- 2}$ , the man's apparent weight on his way down is

1

60 N

2

240 N

3

360 N

4

840 N

5

3600 N

Explanation:

C Given, $m = 60 kg, a_{0} = 4 m / s^{2}$
With respect to the frame of reference of elevator,

$\sum F = 0$
$N + {ma}_{0} = mg$
$N = m (g - a_{0})$
$N = 60 (10 - 4) = 60 \times 6$
$N = 360 N$

Kerala CEE - 2017

LAWS OF MOTION (ADDITIONAL)

371771 A man weighing $100 kg$ slides down a light rope with an acceleration of $1.8 {ms}^{- 2}$ . If $g = 9.8 {ms}^{- 2}$ , the tension of the rope is

1

180 N

2

1160 N

3

800 N

4 weightlessness

Explanation:

C Given, mass of man $(m) = 100 kg$ acceleration $= 1.8 m / s^{2}$

We know that,
$∴ T + ma = mg$
$T = m (g - a)$
$T = 100 (9.8 - 1.8) = 800 N$
Hence, the tension of the rope is $800 N$

Manipal UGET-2017

LAWS OF MOTION (ADDITIONAL)

371772 A body of mass $6 kg$ is hanging from another body of mass $10 kg$ as shown in figure. This combination is being pulled up by a string with an acceleration of $2 {ms}^{- 2}$ . The tension $T_{1}$ is, $(g =$ $10 {m s}^{- 2}$ )

1

240 N

2

150 N

3

220 N

4

192 N

5

178 N

Explanation:

D Given,
$m_{1} = 10 kg$
$m_{2} = 6 kg$
mass of system, $M = m_{1} + m_{2}$

$T_{1} = M (g + a)$
Tension is pulling the mass $(M = 10 + 6 = 16 kg)$
$T_{1} = (10 + 6) (10 + 2)$
$T_{1} = 192 N$

Kerala CEE - 2010

LAWS OF MOTION (ADDITIONAL)

371773 An object of mass $5 kg$ is attached to the hook of a spring balance and the balance is suspended vertically from the roof of a lift. The reading on the spring balance when the lift is going up with an acceleration of $0.25 {ms}^{- 2}$ is taken $(g = 10 {ms}^{- 2})$

1

51.25 N

2

48.75 N

3

52.75 N

4

47.25 N

5

55 N

Explanation:

A Given that,
$m = 5 kg$
$a = 0.25 m / s^{2}$

Let $N$ be the reading on spring Balance-

Force will be equal to the weight-
Weight $= m g + m a$
$= m (g + a)$
Weight $= 5 (10 + 0.25)$
$∴ = 51.25 N$

Kerala CEE - 2008

LAWS OF MOTION (ADDITIONAL)

371770 A man of mass $60 kg$ climbed down using an elevator. The elevator had an acceleration 4 ${ms}^{- 2}$ . If the acceleration due to gravity is $10$ ${ms}^{- 2}$ , the man's apparent weight on his way down is

1

60 N

2

240 N

3

360 N

4

840 N

5

3600 N

Explanation:

C Given, $m = 60 kg, a_{0} = 4 m / s^{2}$
With respect to the frame of reference of elevator,

$\sum F = 0$
$N + {ma}_{0} = mg$
$N = m (g - a_{0})$
$N = 60 (10 - 4) = 60 \times 6$
$N = 360 N$

Kerala CEE - 2017

LAWS OF MOTION (ADDITIONAL)

371771 A man weighing $100 kg$ slides down a light rope with an acceleration of $1.8 {ms}^{- 2}$ . If $g = 9.8 {ms}^{- 2}$ , the tension of the rope is

1

180 N

2

1160 N

3

800 N

4 weightlessness

Explanation:

C Given, mass of man $(m) = 100 kg$ acceleration $= 1.8 m / s^{2}$

We know that,
$∴ T + ma = mg$
$T = m (g - a)$
$T = 100 (9.8 - 1.8) = 800 N$
Hence, the tension of the rope is $800 N$

Manipal UGET-2017

LAWS OF MOTION (ADDITIONAL)

371772 A body of mass $6 kg$ is hanging from another body of mass $10 kg$ as shown in figure. This combination is being pulled up by a string with an acceleration of $2 {ms}^{- 2}$ . The tension $T_{1}$ is, $(g =$ $10 {m s}^{- 2}$ )

1

240 N

2

150 N

3

220 N

4

192 N

5

178 N

Explanation:

D Given,
$m_{1} = 10 kg$
$m_{2} = 6 kg$
mass of system, $M = m_{1} + m_{2}$

$T_{1} = M (g + a)$
Tension is pulling the mass $(M = 10 + 6 = 16 kg)$
$T_{1} = (10 + 6) (10 + 2)$
$T_{1} = 192 N$

Kerala CEE - 2010

LAWS OF MOTION (ADDITIONAL)

371773 An object of mass $5 kg$ is attached to the hook of a spring balance and the balance is suspended vertically from the roof of a lift. The reading on the spring balance when the lift is going up with an acceleration of $0.25 {ms}^{- 2}$ is taken $(g = 10 {ms}^{- 2})$

1

51.25 N

2

48.75 N

3

52.75 N

4

47.25 N

5

55 N

Explanation:

A Given that,
$m = 5 kg$
$a = 0.25 m / s^{2}$

Let $N$ be the reading on spring Balance-

Force will be equal to the weight-
Weight $= m g + m a$
$= m (g + a)$
Weight $= 5 (10 + 0.25)$
$∴ = 51.25 N$

Kerala CEE - 2008

LAWS OF MOTION (ADDITIONAL)

371770 A man of mass $60 kg$ climbed down using an elevator. The elevator had an acceleration 4 ${ms}^{- 2}$ . If the acceleration due to gravity is $10$ ${ms}^{- 2}$ , the man's apparent weight on his way down is

1

60 N

2

240 N

3

360 N

4

840 N

5

3600 N

Explanation:

C Given, $m = 60 kg, a_{0} = 4 m / s^{2}$
With respect to the frame of reference of elevator,

$\sum F = 0$
$N + {ma}_{0} = mg$
$N = m (g - a_{0})$
$N = 60 (10 - 4) = 60 \times 6$
$N = 360 N$

Kerala CEE - 2017

LAWS OF MOTION (ADDITIONAL)

371771 A man weighing $100 kg$ slides down a light rope with an acceleration of $1.8 {ms}^{- 2}$ . If $g = 9.8 {ms}^{- 2}$ , the tension of the rope is

1

180 N

2

1160 N

3

800 N

4 weightlessness

Explanation:

C Given, mass of man $(m) = 100 kg$ acceleration $= 1.8 m / s^{2}$

We know that,
$∴ T + ma = mg$
$T = m (g - a)$
$T = 100 (9.8 - 1.8) = 800 N$
Hence, the tension of the rope is $800 N$

Manipal UGET-2017

LAWS OF MOTION (ADDITIONAL)

371772 A body of mass $6 kg$ is hanging from another body of mass $10 kg$ as shown in figure. This combination is being pulled up by a string with an acceleration of $2 {ms}^{- 2}$ . The tension $T_{1}$ is, $(g =$ $10 {m s}^{- 2}$ )

1

240 N

2

150 N

3

220 N

4

192 N

5

178 N

Explanation:

D Given,
$m_{1} = 10 kg$
$m_{2} = 6 kg$
mass of system, $M = m_{1} + m_{2}$

$T_{1} = M (g + a)$
Tension is pulling the mass $(M = 10 + 6 = 16 kg)$
$T_{1} = (10 + 6) (10 + 2)$
$T_{1} = 192 N$

Kerala CEE - 2010

LAWS OF MOTION (ADDITIONAL)

371773 An object of mass $5 kg$ is attached to the hook of a spring balance and the balance is suspended vertically from the roof of a lift. The reading on the spring balance when the lift is going up with an acceleration of $0.25 {ms}^{- 2}$ is taken $(g = 10 {ms}^{- 2})$

1

51.25 N

2

48.75 N

3

52.75 N

4

47.25 N

5

55 N

Explanation:

A Given that,
$m = 5 kg$
$a = 0.25 m / s^{2}$

Let $N$ be the reading on spring Balance-

Force will be equal to the weight-
Weight $= m g + m a$
$= m (g + a)$
Weight $= 5 (10 + 0.25)$
$∴ = 51.25 N$

Kerala CEE - 2008