Question

A 1.8 Kg block is moved at constant speed over a surface for which the coefficient of friction is 1/4. It is pulled by a force F acting at 45° with horizontal. The block is displaced by 2 m. Find the work done by :
a) Force F

b) Friction

c) Gravity


Refer to the attachment


Thank You​

Answer 1

Explanation:

$\huge\blue{\underline{\underline{\bf Question:-}}}$

A 1.8 Kg block is moved at constant speed over a surface for which the coefficient of friction is 1/4. It is pulled by a force F acting at 45° with horizontal. The block is displaced by 2 m. Find the work done by :

a) Force F

b) Friction

c) Gravity

$\huge\blue{\underline{\underline{\bf Answer:-}}}$

By newtons second law of motion

$f_{net} = ma$

$N + F sin 45 = mg \\ \\ N = 1.8 \times 10 - \frac{F}{ \sqrt{2} } \\ \\ \boxed{ N = 18 - \frac{F}{ \sqrt{2} } ...........(1)}$

Now,

$F sin 45 - \mu N = 0 \\ \\ \frac{F}{ \sqrt{2} } = \frac{1}{4} (18 - \frac{ F}{ \sqrt{2} } ) \\ \\ \frac{ F}{ \sqrt{2} } = \frac{18}{4} - \frac{ F}{4 \sqrt{2} } \\ \\ \frac{5 F}{ \cancel{4} \sqrt{2} } = \frac{18}{ \cancel4} \\ \\ \boxed{ F = \frac{18 \sqrt{2} }{5} }$

Now,

Work done by force will be

$W_{F} = F sin 45 \times S \\ \\ W_{F } = \frac{18 \cancel{ \sqrt{2}} }{5} \times \frac{1}{ \cancel{\sqrt{2}} } \times 2 \\ \\ \boxed{ W_{F} = \frac{36}{5} = 7.2}$

Now let's calculate Normal using (1)

$N = 18 - \frac{F }{ \sqrt{2} } \\ \\ N = 18 - \frac{18 \cancel{ \sqrt{2} }}{5 \cancel{\sqrt{2} }} \\ \\ \boxed{ N = \frac{72}{5} = 14.4} \\ \\ \\ \\ W_{friction} = \mu N S \\ \\ W_{friction} = - \frac{1}{ \cancel4} \times \frac{72}{5} \times \cancel2 \\ \\ \boxed{W_{friction} = - \frac{36}{5} } \\ \\ \\ \\ W_{gravity} = mg \times S \times \cos(90) \\ \\ W_{gravity} = mg \times 2 \times 0 \\ \\ \boxed{W_{gravity} = 0}$

$ANSWERS:- \\ \\ a) \frac{36}{5} \\ \\ b) - \frac{36}{5} \\ \\ c ) 0$

Answer 2

${ \huge \bf { \mid{ \overline{ \underline{Question}}} \mid}}$

A 1.8 Kg block is moved at constant speed over a surface for which the coefficient of friction is 1/4. It is pulled by a force F acting at 45° with horizontal. The block is displaced by 2 m. Find the work done by

a)Force "F"

b)Friction

c)Gravity.

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

• Displacement (s) = 2m.
• Mass of the Block (m) = 1.8 kg.
• Angle Between Force & Displacement (θ) = 45°.
• Coffecient of Friction (μ) = 1/4 = 0.25

Assumptions:-

• Frictional Force = ${F_f}$
• Force due to Gravity (weight) = ${F_G}$
• Force applied on the Block = F.
• Normal reaction = ${F_N}$

$\huge{\bold{\underline{Explanation:-}}}$

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Applying Work done formula,

$\large{\boxed{\tt W = F.s cos \theta}}$

Substituting the values,

$\large{\tt \leadsto W = F \times 2 \times cos 45 \degree}$

$\large{\tt \leadsto W = 2F \times \dfrac{1}{\sqrt{2}}}$

$\large{\tt \leadsto W = \cancel{2}F \times \dfrac{1}{\cancel{\sqrt{2}}}}$

$\large{\leadsto {\underline{\underline{\tt W = \sqrt{2}F}}} \: \tt ------(1)}$

Now, Finding the Value of Force.

Dividing the Force "F" into Components.

I.e.

• Fcosθ ( Along Horizontal) &
• Fsinθ ( Along Vertical)

CASE-1

From Equilibrium of Particle,

$\large{\underline{\underline{\tt F_N + F sin 45 = mg}}}$

As F sin 45 is Acting vertically upwards,

$\large{\tt \leadsto F_N + F sin 45 = mg}$

$\large{\leadsto {\underline{\tt F_N = mg - F sin 45}}}$

Note:-

• Here Normal reaction ≠ weight (mg) of the body.

CASE-2

From Equilibrium of particle,

$\large{\underline{\underline{\tt F_f = F cos 45}}}$

As F cos 45 is Acting Horizontally,

And We know,

Frictional Force = Normal reaction × Coffecient of Friction

Substituting,

$\large{\tt \leadsto \mu F_N = F cos 45}$

Substituting the values,

$\large{\tt \leadsto \dfrac{1}{4} \times [mg - F sin 45] = F cos 45}$

$\large{\tt \leadsto \dfrac{mg}{4} - \dfrac{F sin 45}{4} = F cos 45}$

$\large{\tt \leadsto F cos 45 + \dfrac{F sin 45}{4} = \dfrac{mg}{4}}$

cos 45 = sin 45 = 1/√2,

Substituting

$\large{\tt \leadsto \dfrac{F}{\sqrt{2}} + \dfrac{F}{4 \sqrt{2}} = \dfrac{1.8 \times 10}{4}}$

$\large{\tt \leadsto \dfrac{4F + F}{4 \sqrt{2}} = \dfrac{18}{4}}$

$\large{\tt \leadsto \dfrac{5F}{4 \sqrt{2}} = \dfrac{18}{4}}$

$\large{\tt \leadsto \dfrac{5F}{\cancel{4} \sqrt{2}} = \dfrac{18}{\cancel{4}}}$

$\large{\tt \leadsto \dfrac{5F}{\sqrt{2}} = 18}$

$\large{\boxed{\tt F = \dfrac{18 \sqrt{2}}{5}}}$

Now, Substituting the value of Force in Equation (1).

$\large{\leadsto {\underline{\underline{\tt W = \sqrt{2}F}}} \: \tt ------(1)}$

$\large{\tt \leadsto W = \sqrt{2} \times \dfrac{18 \sqrt{2}}{5}}$

$\large{\tt \leadsto W = \dfrac{18 \times 2}{5}}$

$\large{\tt \leadsto W = \dfrac{36}{5}}$

$\huge{\boxed{\boxed{\tt W = 7.2 \: J}}}$

So, the Work done by the Force is 7.2 Joules.

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Applying Work done formula,

$\large{\boxed{\tt W = F.s cos \theta}}$

$\large{\tt \leadsto W = F_f .s cos \theta}$

$\large{\tt \leadsto W = ( \mu \times F_N) . s cos \theta}$

From Above we Know,

$\large{\underline{\tt \mu \times F_N = F cos 45}}$

Substituting it,

$\large{\tt \leadsto W =( F cos 45) . s cos \theta}$

Here θ will be 180° as it's opposite in the Direction of motion.

$\large{\tt \leadsto W = \dfrac{F}{\sqrt{2}} \times 2 \times cos 180}$

$\large{\tt \leadsto W = \dfrac{F}{\cancel{\sqrt{2}}} \times \cancel{2} \times - 1}$

$\large{\tt \leadsto W = \sqrt{2} F \times - 1}$

$\large{\tt \leadsto W = - F \sqrt{2}}$

Substituting the value of Force,

$\large{\tt \leadsto W = - \sqrt{2} \times \dfrac{18 \sqrt{2}}{5}}$

$\large{\tt \leadsto W = \dfrac{- 18 \times 2}{5}}$

$\large{\tt \leadsto W = \dfrac{- 36}{5}}$

$\huge{\boxed{\boxed{\tt W = - 7.2 \: J}}}$

So, the Work done by the Friction is - 7.2 Joules.

Note:-

• Here Negative sign indicates that the work done by the friction is opposing the motion of the body.

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Applying Work done formula,

$\large{\boxed{\tt W = F.s cos \theta}}$

$\large{\tt \leadsto W = F_G .s cos \theta}$

Here θ = 90°

Substituting, we get

$\large{\tt \leadsto W = F_G .s \times cos 90}$

$\large{\tt \leadsto W = F_G .s \times 0}$

$\huge{\boxed{\boxed{\tt W = 0 \: J}}}$

So, the Work done by the Gravity is 0 Joules.

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