Q. 93.8( 8 Votes )

A block of mass 2

Class 11th HC Verma - Concepts of Physics Part 1 8. Work and Energy

Answer :

Given mass m= 0.25kg, inclination θ =37° , displacement = 1m Now weight = mg= 0.25× 10=2.5N (g=10m/)

Now from the F.B.D we find that the force against is the component of weight along the incline = So work done against the friction force on the block = F× s

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PREVIOUSA force F = a + b NEXTA block of mass m

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