Question

A 5 kg block is placed over the rough surface of a
trolley of mass 45 kg which stands on a smooth
horizontal surface. A horizontal force of 5 N is
applied on the block. The minimum coefficient of
friction required to prevent slipping between the block
and trolley is (g = 10 m/s2)
(1) 1
(2) 0.09
(3) 0.9
(4) 0.1​

Answer 1

Answer:

(4) 0.1

Explanation:

force req = normal force * coefficient of friction .

5 = 50 * coefficient

coefficient = 0.1

Answer 2

The minimum coefficient of  friction is 0.1.

(4) is correct option.

Explanation:

Given that,

Mass of block = 5 kg

Mass of trolley = 45 kg

Force = 5 N

We need to calculate the coefficient of  friction

Using balance equation

$\text{Horizontal force} = \text{coefficient of friction}\times mg$

$F_{h}=\mu\times mg$

Where, m = mass of block

F = horizontal force

g = acceleration due to gravity

Put the value into the formula

$5=\mu\times 5\times10$

$\mu=\dfrac{5}{50}$

$\mu=\dfrac{1}{10}$

$\mu=0.1$

Hence, The minimum coefficient of  friction is 0.1.

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Topic : coefficient of  friction

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