Question

A 30 kg block rests on a rough horizontal surface. A force of 200 n is applied on the block. Tge block acquires a speed of 4 m/s starting from rest in 2s . What is tge value of coefficient of friction

Answer 1

Answer:

The answer will be 0.47

Explanation:

According to the problem the mass of the block is 30 kg .

The applied force on  the block is 200 n and the speed requires to move is 4 m/s

Now the the initial velocity of the block , u = 0

after 2 sec the velocity of the block is v = 4 m/s

As the force is applied to move the block a force is also applied by the ground in the opposite direction  as friction

friction f = μx n

Now the acceleration of the block F = ma

                                                     => 200-f = 30a .....(1)

Now v = u +at

      => 4 = 0 +2a

       => a = 2m/s^2

Now putting it on 1

200 -f = 30 x 2

=> f = 140

Therefore f = μx n

normal,n  = mg

f = μx mg

=> 140 = μx 30 x 10

=> μ = 0.47

Answer 2

Given :

• m = 30 kg
• F = 200 N
• v = 4 m/s
• u = 0 m/s
• t = 2 s

To Find :

• Value of coefficient of friction

Solution :

By using

$\Large \underline{ \tt v = u + at }\\ \\ \implies \tt4 = 0 + a \times 2 \\ \\ \implies \tt a = \frac{4}{2} \\ \\ \implies \boxed{\tt a = 2 \: {ms}^{ - 2}}$

Force producing acceleration

$\Large \underline{\tt f =ma }\\ \\\tt\implies f = 30 \times 2 \\ \\ \implies \boxed{ \tt f =60 \: N}$

Force of friction

$\Large\tt \underline{\mu R = F - f }\\ \\\implies \tt \mu R = 200 - 60 \\ \\ \implies \tt \mu R = 140 \\ \\\implies \tt \mu = \frac{140}{R} \\ \\ \implies \tt \mu = \frac{140}{mg} \\ \\ \implies\tt \mu = \frac{140}{30 \times 10} \\ \\ \Large \implies \boxed{\boxed{ \tt \green{ \mu = 0.47}}}$