Question

A 6Kg block by a distance of s=12m along a horizontal surface at a constant speed. If the coefficient of kinetic friction is mu=0.2 and the cord pulling the block is at an angle of 45 degree with the horizontal, then calculate the work done by the person. (take g=10m/s2)

Answer= 120 Joule​

Answer 1

Answer:-

Given that :-

• $\sf s = 12\:m$
• $\sf \mu_k = 0.2$
• Angle between applied force and horizontal = angle between applied force and displacement = $\sf \theta = 45^o$

Refer to the attachment for F.B.D.

Since there is no acceleration in vertical direction,

$\sf mg - F\: sin(\theta) = N\quad\quad\dots(1)$

And,

Since it is given, "... along a horizontal surface at a constant speed..." there will be no acceleration in horizontal direction also. So we can say,

$\sf f = F\: cos(\theta)$

And as we know that, $\sf f = \mu_kN$ (kinetic friction is considered since the block is in motion),

$\sf f = F\: cos(\theta) = \mu_kN$

From (1),

$\longrightarrow \sf F\: cos(\theta) = \mu_k\bigg\{ mg - F\:sin(\theta) \bigg\}$

$\longrightarrow \sf F\: cos(\theta) = \mu_kmg - \mu_k F \: sin(\theta)$

$\longrightarrow \sf F\:cos(\theta) + \mu_kF\:sin(\theta) = \mu_kmg$

$\longrightarrow \sf F\{cos(\theta) + \mu_ksin(\theta)\} = \mu_kmg$

$\longrightarrow \sf F = \dfrac{\mu_kmg}{\{cos(\theta) + \mu_ksin(\theta)\}}$

Putting the values given,

$\longrightarrow \sf F = \dfrac{0.2 \times 6 \times 10}{\Bigg\{\bigg(\dfrac{1}{\sqrt{2}}\bigg) + \bigg(\dfrac{0.2}{\sqrt{2}}\bigg) \Bigg\}}$

$\longrightarrow \sf F = \dfrac{12}{1.2}\times \sqrt{2}$

$\longrightarrow \sf F = 10\sqrt{2}\:N$

Now that we have the value of F, we can find the work done by F.

$\sf W_F = F\times s \times cos(\theta)$

$\sf \longrightarrow W_F = 10\sqrt{2} \times 12 \times \dfrac{1}{\sqrt{2}}$

$\longrightarrow \underline{\underline{\sf{\green{W_F = 120\:J}}}}$