If is the force required to just move the block up the inclined plane and is the force required to just prevent the block from sliding down, then the value of is [Use g = 10 ]

Question

A block of mass 5 kg is placed on a rough inclined surface as shown in the figure.

If ${\vec{F}}_{1}$ is the force required to just move the block up the inclined plane and ${\vec{F}}_{2}$ is the force required to just prevent the block from sliding down, then the value of $| {\vec{F}}_{1} |$ $-$ $| {\vec{F}}_{2} |$ is [Use g = 10 $m / s^{2}$ ] [2024]

Smart Achievers · Accepted Answer

(2)

${\vec{F}}_{1}$

$| {\vec{F}}_{1} |$

$F_{1} = m g \sin θ + f_{K} = 25 + 2.5 \sqrt{3}$

$F_{2} = m g \sin θ - f_{K} = 25 - 2.5 \sqrt{3}$

$∴$ $F_{1} - F_{2} = 5 \sqrt{3} N$

Solution

1 $25 \sqrt{3}$ N

2 $5 \sqrt{3}$ N

3 $\frac{5 \sqrt{3}}{2}$ N

4 10 N

Ans.
(2)

$f_{K} = μ m g \cos θ$

$= 0.1 \times \frac{50 \times \sqrt{3}}{2} = 2.5 \sqrt{3} N$

$F_{1} = m g \sin θ + f_{K} = 25 + 2.5 \sqrt{3}$

$F_{2} = m g \sin θ - f_{K} = 25 - 2.5 \sqrt{3}$

$∴$ $F_{1} - F_{2} = 5 \sqrt{3} N$

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Solution

1 253 N

2 53 N

3 532 N