Question

A slab of mass 5 kg lies on a smooth horizontal surface a block of mass 4 kg on the top of it . coefficient of friction between the block and slab is 0.25. If the block is pulled horizontally by a force of F= 6N, what is the work done by the force of friction on the slab between the instance t = 2S to t=3S?(take g=10m/s²)

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Answer 1

Explanation:

ANSWER

Maximum frictional force between the slab and the block

f

max

=μN=μmg=

4

1

×4×10=10N

Evidently, f∝f

max

So, the two bodies will move together as a single unit. If a be their combined acceleration, then

a=

m+M

F

=

4+5

6

=

3

2

ms

−2

Therefore, frictional force acting can be obtained as

f=Ma=

3

2

×5=

3

10

N

Using S=

2

1

at

2

S(2)=

2

1

×

3

2

(2)

2

=

3

4

and S(3)=

2

1

×

3

2

×(3)

2

=3

Therefore, work done by friction =F[S(3)−S(2)]

=

3

10

[3−

3

4

]=

9

50

=5.55J

solution

this is your answer mate hope it helps

Answer 2

Answer:

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