Question

88. A block of mass 5 kg is kept against an
accelerating wedge with a wedge angle of 45°
to the horizontal. The coefficient of friction
between the block and the wedge is u = 0.4.
What is the minimum absolute value of the
acceleration of the wedge to keep the block
steady? [Assume, g = 10 m/s2)

Answer 1

Given that,

Mass = 5 kg

Angle = 45°

Coefficient of friction = 0.4

We need to calculate the maximum force

Using formula of maximum force

$f_{max}=\mu mg\cos\theta$

Where, m = mass

g = acceleration due to gravity

$\mu$ = coefficient of friction

Put the value into the formula

$f_{max}=0.4\times4\times10\cos45^{\circ}$

$f_{max}=11.3\ N$

We need to calculate the absolute value of acceleration

Using balance equation

$mg\sin\theta-\mu mg\cos\theta=ma$

$a=g(\sin\theta-\mu\cos\theta)$

Put the value into the formula

$a=10(\sin45-0.4\times\cos45)$

$a=4.2\ m/s^2$

$a=\dfrac{30}{7}\ m/s^2$

Hence, The absolute value of acceleration is $\dfrac{30}{7}\ m/s^2$

(a) is correct option.

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Topic : acceleration

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