Question

a block of mass 2 kg is placed on surface of trolley of mass 20 kgs which is on a smooth surface. the coefficient of friction between block and surface of trolley is 0.25. if horizontal of 2 N acts on block the acceleration of system

Answer 1

Answer:

Calculating Frictional force between Block and Surface of trolley.

We get,

N = Weight of Block = 2 × 10 = 20 N

Coefficient of friction = 0.25

⇒ Frictional Force = 20 × 0.25 = 5 N

Hence the block won't move on applying 2 N.

Hence both the trolley and block will move as a system.

We know that,

Acceleration = Sum of all forces in a system / Sum of all masses

⇒ Acceleration = 2 / ( 2 + 20 )

⇒ Acceleration = 2 / 22 = 1 / 11 = 0.090 m/s²

This is the required answer

Hope it helps :)

Answer 2

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N = Weight of Block

= 2 × 10 = 20 N

Coefficient of friction = 0.25

⇒ Frictional Force = 20 × 0.25

= 5 N

Here

Block won't move on applying 2 N.

Hence

Both the trolley and block will move as a sequence.

We know that,

Acceleration = $\huge\bf \frac{Sum of all forces in a system}{Sum of all masses }$

⇒ Acceleration = $\huge\bf \frac{2}{2 + 20}$

Acceleration

= $\huge\bf \frac{2}{22}$

=  $\huge\bf \frac{1}{11}$

= 0.090 $m/s^{2}$

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