Question

A block of mass 5kg is pulled with a constant force of 25N with the help of a
string. The coefficient of friction between the block and the ground is 0.2. Pulley is
massless and string is massless & inextensible. When the string makes an angle of
37" with the horizontal, the acceleration of the block is
(A) 2
(B4m/s
(C)5m/s
(D) 2.6m​

Answer 1

Answer:

the answer = 2m/s

let acceleration is friction M=0.2

F=25N

ma=Fcos 37-mgM

therefore;a=2m/s

Answer 2

The acceleration of the block is (a) 2 m/s²

Given: Mass of block = 5 kg, Magnitude of force = 25 N,

The coefficient of friction between the block and the ground is 0.2.

The string makes an angle of 37° with the horizontal.

To Find: The acceleration of the block.

Solution:

• The question can be solved by using the formula,

            ∑ F = m × a             [ F = force, m = mass, a = acceleration ]

• We know that frictional force always acts in the direction opposite to the direction of motion.

Coming to the numerical,

m = 5 kg, F = 25 N, μ = 0.2,

Component of force acting along the direction of force = F cos 37°

                                                                                            = 25 × 4 / 5

                                                                                            = 20 N

Force of friction = μmg

                          = 0.2 × 5 × 10

                          = 10 N

According to the question,

             ∑ F = m × a    

     ⇒  F cos 37° - μmg = m × a

     ⇒  20 - 10 = 5 × a

     ⇒     a = 10 / 5

     ⇒     a = 2 m/s²

Hence, the acceleration of the block is (a) 2 m/s².

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