Question

Q.A 10 g block placed on a rough horizontal floor friction between the block and the floor is 0.4. Find work done
& being pulled by a constant force 50 N. Coefficient of kinetic by each individual force acting on the block over displacement of 5 m.

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Answer 1

Given :

Mass of the block = 10g = 0.01kg

Coefficient of kinetic friction = 0.4

Applied force = 50N

To Find :

Work done by each individual force acting on the force over displacement of 5m.

Solution :

★ Work done is measured as the dot product of force and displacement.

• It is a scalar quantity having only magnitude.
• SI unit : Joule (J)
• Dimension formula : [M¹L²T‾²]

Mathematically,

$\dag\:\underline{\boxed{\bf{\orange{W=\overrightarrow{F}\cdot\overrightarrow{d}=Fd\:cos\theta}}}}$

Where θ denotes angle between force and displacement vectors.

A] Work done by external force :

$\sf:\implies\:W=F_ed\:cos\theta$

$\sf:\implies\:W_1=(50)(5)\:cos0^{\circ}$

$\sf:\implies\:W_1=250\times 1$

$\bf:\implies\:W_1=250\:J$

B] Work done by frictional force :

$\sf:\implies\:W_2=F_fd\:cos\theta$

$\sf:\implies\:W_2=(\mu_kN)d\:cos\theta$

$\sf:\implies\:W_2=(\mu_k\times mg)d\:cos180^{\circ}$

$\sf:\implies\:W_2=0.4\times0.01\times10\times5\times (-1)$

$\sf:\implies\:W_1=0.2\times (-1)$

$\bf:\implies\:W_2=-0.2\:J$

C] Work done by net force :

➠ W = W₁ + W₂

➠ W = 250 + (-0.2)

➠ W = 249.8 N

D] Work done by weight force :

$\sf:\implies\:W_3=F_wd\:cos\theta$

$\sf:\implies\:W_3=(mg)d\:cos\theta$

$\sf:\implies\:W_3=(0.01)(10)(5)\:cos90^{\circ}$

$\bf:\implies\:W_3=0\:J$

Answer 2

Solution

Net force acting on the block

F - F = ma

50 - ųmg = ma

50 - 0.44 × 10 × 10 = ma

50 - 40 = ma

$\sf \: a = \dfrac{10}{10} = 1m \: / {s}^{2}$

Net force = 50 - 40 = 10 N

Work done by constant force = F × displacement

50 × 5 = 250 J

Work done by friction :

= -F × 5

= -40 × 5 = -200J

It is negative Since displacement is opposite of the direction of friction

Net work = 250 - 200 = 50J