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Laws of Motion

270333 A man slides down on a telegraphic pole with an acceleration equal to one-fourth of acceleration due to gravity. The frictional force between man and pole is equal to (in terms of man's weight $W$ )

1

\frac{W}{4}

2

\frac{3 W}{4}

3

\frac{W}{2}

4

W

Explanation:

$f = w_{a p p} = m (g - a)$

Laws of Motion

270334 A box is placed on the floor of a truck moving with an acceleration of $7 {ms}^{- 2}$ . If the coefficient of kinetic friction between the box and surface of the truck is 0.5 ,find the acceleration of the box relative to the truck

1

1.7 {ms}^{- 2}

2

2.1 {ms}^{- 2}

3

3.5 {ms}^{- 2}

4

4.5 {ms}^{- 2}

Explanation:

$f_{k} = μ_{k} N, N = m g; a^{1} = μ_{k} g, a_{r e l} = a - a^{1}$

Laws of Motion

270335 A block is placed at a distance of $2 m$ from the rear on the floor of a truck $(g = 10 {ms}^{- 2})$ . When the truck moves with an acceleration of $8 {ms}^{- 2}$ , the block takes $2 \sec$ to fall off from the rear of the truck. The coefficient of sliding friction between truck and the block is

1 0.5

2 0.1

3 0.8

4 0.7

Explanation:

$s = u t + \frac{1}{2} a t^{2}, a^{1} = μ_{k} g, a_{r e l} = a - a^{1}$

Laws of Motion

270336 Sand is piled up on a horizontal ground in the form of a regular cone of a fixed base of radius $R$ . The coefficient of static friction between sand layers is $μ$ . The maximum volume of sand that can be piled up, without the sand slipping on the surface is

1

\frac{μ R^{3}}{3 π}

2

\frac{μ R^{3}}{3}

3

\frac{π R^{3}}{3 μ}

4

\frac{μ π R^{3}}{3}

Explanation:

$\tan θ = \frac{h}{R} = μ; V = \frac{1}{3} π R^{2} h$

Laws of Motion

270333 A man slides down on a telegraphic pole with an acceleration equal to one-fourth of acceleration due to gravity. The frictional force between man and pole is equal to (in terms of man's weight $W$ )

1

\frac{W}{4}

2

\frac{3 W}{4}

3

\frac{W}{2}

4

W

Explanation:

$f = w_{a p p} = m (g - a)$

Laws of Motion

270334 A box is placed on the floor of a truck moving with an acceleration of $7 {ms}^{- 2}$ . If the coefficient of kinetic friction between the box and surface of the truck is 0.5 ,find the acceleration of the box relative to the truck

1

1.7 {ms}^{- 2}

2

2.1 {ms}^{- 2}

3

3.5 {ms}^{- 2}

4

4.5 {ms}^{- 2}

Explanation:

$f_{k} = μ_{k} N, N = m g; a^{1} = μ_{k} g, a_{r e l} = a - a^{1}$

Laws of Motion

270335 A block is placed at a distance of $2 m$ from the rear on the floor of a truck $(g = 10 {ms}^{- 2})$ . When the truck moves with an acceleration of $8 {ms}^{- 2}$ , the block takes $2 \sec$ to fall off from the rear of the truck. The coefficient of sliding friction between truck and the block is

1 0.5

2 0.1

3 0.8

4 0.7

Explanation:

$s = u t + \frac{1}{2} a t^{2}, a^{1} = μ_{k} g, a_{r e l} = a - a^{1}$

Laws of Motion

270336 Sand is piled up on a horizontal ground in the form of a regular cone of a fixed base of radius $R$ . The coefficient of static friction between sand layers is $μ$ . The maximum volume of sand that can be piled up, without the sand slipping on the surface is

1

\frac{μ R^{3}}{3 π}

2

\frac{μ R^{3}}{3}

3

\frac{π R^{3}}{3 μ}

4

\frac{μ π R^{3}}{3}

Explanation:

$\tan θ = \frac{h}{R} = μ; V = \frac{1}{3} π R^{2} h$

Laws of Motion

270333 A man slides down on a telegraphic pole with an acceleration equal to one-fourth of acceleration due to gravity. The frictional force between man and pole is equal to (in terms of man's weight $W$ )

1

\frac{W}{4}

2

\frac{3 W}{4}

3

\frac{W}{2}

4

W

Explanation:

$f = w_{a p p} = m (g - a)$

Laws of Motion

270334 A box is placed on the floor of a truck moving with an acceleration of $7 {ms}^{- 2}$ . If the coefficient of kinetic friction between the box and surface of the truck is 0.5 ,find the acceleration of the box relative to the truck

1

1.7 {ms}^{- 2}

2

2.1 {ms}^{- 2}

3

3.5 {ms}^{- 2}

4

4.5 {ms}^{- 2}

Explanation:

$f_{k} = μ_{k} N, N = m g; a^{1} = μ_{k} g, a_{r e l} = a - a^{1}$

Laws of Motion

270335 A block is placed at a distance of $2 m$ from the rear on the floor of a truck $(g = 10 {ms}^{- 2})$ . When the truck moves with an acceleration of $8 {ms}^{- 2}$ , the block takes $2 \sec$ to fall off from the rear of the truck. The coefficient of sliding friction between truck and the block is

1 0.5

2 0.1

3 0.8

4 0.7

Explanation:

$s = u t + \frac{1}{2} a t^{2}, a^{1} = μ_{k} g, a_{r e l} = a - a^{1}$

Laws of Motion

270336 Sand is piled up on a horizontal ground in the form of a regular cone of a fixed base of radius $R$ . The coefficient of static friction between sand layers is $μ$ . The maximum volume of sand that can be piled up, without the sand slipping on the surface is

1

\frac{μ R^{3}}{3 π}

2

\frac{μ R^{3}}{3}

3

\frac{π R^{3}}{3 μ}

4

\frac{μ π R^{3}}{3}

Explanation:

$\tan θ = \frac{h}{R} = μ; V = \frac{1}{3} π R^{2} h$

Laws of Motion

270333 A man slides down on a telegraphic pole with an acceleration equal to one-fourth of acceleration due to gravity. The frictional force between man and pole is equal to (in terms of man's weight $W$ )

1

\frac{W}{4}

2

\frac{3 W}{4}

3

\frac{W}{2}

4

W

Explanation:

$f = w_{a p p} = m (g - a)$

Laws of Motion

270334 A box is placed on the floor of a truck moving with an acceleration of $7 {ms}^{- 2}$ . If the coefficient of kinetic friction between the box and surface of the truck is 0.5 ,find the acceleration of the box relative to the truck

1

1.7 {ms}^{- 2}

2

2.1 {ms}^{- 2}

3

3.5 {ms}^{- 2}

4

4.5 {ms}^{- 2}

Explanation:

$f_{k} = μ_{k} N, N = m g; a^{1} = μ_{k} g, a_{r e l} = a - a^{1}$

Laws of Motion

270335 A block is placed at a distance of $2 m$ from the rear on the floor of a truck $(g = 10 {ms}^{- 2})$ . When the truck moves with an acceleration of $8 {ms}^{- 2}$ , the block takes $2 \sec$ to fall off from the rear of the truck. The coefficient of sliding friction between truck and the block is

1 0.5

2 0.1

3 0.8

4 0.7

Explanation:

$s = u t + \frac{1}{2} a t^{2}, a^{1} = μ_{k} g, a_{r e l} = a - a^{1}$

Laws of Motion

270336 Sand is piled up on a horizontal ground in the form of a regular cone of a fixed base of radius $R$ . The coefficient of static friction between sand layers is $μ$ . The maximum volume of sand that can be piled up, without the sand slipping on the surface is

1

\frac{μ R^{3}}{3 π}

2

\frac{μ R^{3}}{3}

3

\frac{π R^{3}}{3 μ}

4

\frac{μ π R^{3}}{3}

Explanation:

$\tan θ = \frac{h}{R} = μ; V = \frac{1}{3} π R^{2} h$