Question

A boy pulls a 5 kg block 20 meters along a horizontal surface at a constant speed with a force directed 45 degrees above the horizontal. If the coefficient of kinetic friction is 0.2, how much work does the boy do on the block-

A) 163.32 J B)11.55 J C)150 J D)115 J !!! ( the ques is easy....but my answer comes to nearly 236 which is not in the option ..kindly help )

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Eklavya asked in Physics

A boy pulls a 5 kg block 20 meters along a horizontal surface at a constant speed with a force directed 45 degrees above the horizontal. If the coefficient of kinetic friction is 0.2, how much work does the boy do on the block-

A) 163.32 J B)11.55 J C)150 J D)115 J !!! ( the ques is easy....but my answer comes to nearly 236 which is not in the option ..kindly help )

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Kranav.sharma... answered this
6497 helpful votes in Physics, Class XII-Science

Here,

Mass of the block, m = 5 kg

Coefficient of friction, μ = 0.2

Distance traveled, d = 20 m

Let the force be F. F makes an angle 45o with the horizontal.

The vertical component of the applied force is F sin45.

Thus the net force by the block on the ground = (mg – F sin45)

So, the frictional force acting on the block is = μ(mg – F sin45)

The block moves at a constant speed. That means the horizontal component of the force just nullifies the frictional force.

So, F cos45 = μ(mg – F sin45)

=> F cos45 + μF sin45 = μmg

=> (F cos45)(1 + μ tan45) = μmg

=> (F cos45)(1 + μ) = μmg

=> F cos45 = μmg/(1 + μ)

Now the work done on the block is = (F)(d)(cos45)

= (F cos45)(d)

= [μmg/(1 + μ)](d)

= (0.2)(5)(9.8)(20)/(1+0.2)

= 196/1.2

= 163.33 J

Correct option is (a).

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A K Daya Sir answered this
4894 helpful votes in Physics, Class XI-Science

Let the boy pulls the block with a force F making angle Q with horizontal.

F can have two components, F cosQ along horizontal, and FsinQ along vertical.

FsinQ helps to lower the normal reaction, whereas FcosQ works against friction(uR)

R = mg - FsinQ

FcosQ = uR = 0.2 R

FcosQ = 0.2 (mg - FsinQ)

FcosQ + 0.2FsinQ = 0.2mg

F(cos45 + 0.2 sinQ) = 0.2 mg

F (0.707 + 0.141) = 0.2 x 5 x 9.8

F x 0.848 = 9.8

F = 9.8/0.848 = 11.56 N

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Work done = Fcos45 x 20 = 11.55 x 0.707 x 20 = 163.32 J

option A