Question

a block of weight 150N is pulled 20m along a horizontal surface at constant velocity. calculate the work done by the pulling force if the coefficient of kinetic friction is 0.20 and the pulling force makes an angle of 60 degree with the vertical.
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Answer 1

Answer:

1200 joules

Explanation:

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Answer 2

Given :

The weight of the block = w = 150 N

The displacement along horizontal surface = 20 ,

Coefficient of kinetic friction = $\mu$ = 0.20

The pulling force makes an angle with the vertical = 60°

To Find :

The work done by the pulling force

Solution :

An angle of 60° with the vertical is 30° with the horizontal

Now,

Let The pulling force F provides a force parallel to the surface = $F_p$

And   $F_p$ = F cos 30°

i.e       $F_p$ = F × 0.866

Similarly

Let The same force of F also provides a force normal to the surface =

And  $F_n$ = F sin 30°

i.e       $F_n$ = F × 0.5

∴   The net weight of the block = $F_n_e_t$ = weight of block - $F_n$

i.e    $F_n_e_t$ = ( 150 - 0.5 F ) N

Again

∵   The pulling force F provides a force parallel to the surface = $F_p$ = 0.866 F            ............1

And  , $\mu$ = 0.20

So,  $F_p$  = 0.20 × $F_n_e_t$            

i.e    $F_p$  = 0.20 × ( 150 - 0.5 F )       ...........2

From eq 1 and eq 2

Or,       0.20 × ( 150 - 0.5 F )  = 0.866 F

Or,   30 - 0.1 F = 0.866 F

Or,   0.1 F + 0.866 F = 30

Or,   0.966 F = 30

∴                F = $\dfrac{30}{0.966}$

i.e              F = 31 .06 N

So, The value of pulling force = $F_p$ = 0.866 F = 31.06  × 0.866 N

Or,                                               $F_p$ = 26.89 N

Now,

∵    work done = Force × displacement

so, Work done by the pulling force = W = $F_p$ × displacement

Or,             W = 26.89 N × 20 m

∴     Work done = 537.8    joule

Hence, The work done by the pulling force is 537.8  joule   Answer