Question

A box weighing 2000 N is to be slowly slid through 20 m on a straight track with friction coefficient 0⋅2 with the box. (a) Find the work done by the person pulling the box with a chain at an angle θ with the horizontal. (b) Find the work when the person has chosen a value of θ, which ensures him the minimum magnitude of the force.

Answer 1

(a)The work done by the person is  $\begin{equation}\frac{40000}{(5+\tan \theta)}\end$

(b)The work when the person has chosen a value of θ, which ensures him the minimum magnitude of the force is 7692 .3 J.

Explanation:

Step 1:

The conditions are as follows:-

Here weight of the box = 2000 N

μ = 0.2

distance s = 20 m.

So if the bt pull is used, then the individual is F.  

Step 2:

The Natural force, then,

N + F sinθ = W

N = W - F sinθ

N = 2000 - F sinθ

Step 3:

Force of friction is = μN

Force of friction is = μ (2000 - F sinθ)

Force of friction is = 0.2 (2000-F sinθ)

Force of friction is = 400 - 0.2 F sinθ

The driving force is equivalent to the frictional force in accordance with the given conditions.

so,

F. cosθ = 400 - 0.2 F sinθ

400 = F ( cosθ + 0.2 sinθ)

Multiply by 5 on both sides,

400 × 5 = F (cosθ + 0.2 sinθ)×5

2000 = F(5 cosθ + sinθ)

$\begin{equation}\mathrm{F}=\frac{2000}{(\cos \theta+\sin \theta)}$

(a) Now for work done  

work done by the person pulling the box is

= F cosθ × 20 J

$\begin{equation}=\frac{2000}{(\cos \theta+\sin \theta)}[\mathrm{F} \cos \theta \times 20 \mathrm{J}]\end$

$\begin{equation}\mathrm{w}=\frac{40000}{(5+\tan \theta)}\end$ ……………………eqn(1)

Hence the work done by the person is $\begin{equation}\frac{40000}{(5+\tan \theta)}\end$

(b) For minimising the F,  

$\begin{equation}\frac{d f}{d \theta}=0\end$

$\begin{equation}\frac{-2000(-5 \sin \theta+\cos \theta)}{(5 \cos \theta+\sin \theta)^{2}}=0\end$

Hence

cosθ = 5 sinθ

$\begin{equation}\frac{\sin \theta}{\cos \theta}=\frac{1}{5}\end$

tanθ = $\frac{1}{5}$

Therefore, by placing this value in equation I the work has been done in expression.

$\begin{equation}\begin{aligned}&\begin{aligned}W &=\frac{40000}{\left(5+\frac{1}{5}\right) \mathrm{J}} \\W &=\frac{40000}{\left(\frac{26}{5}\right) \mathrm{J}}\end{aligned}\\&W=\frac{40000 \times 5}{(26 \mathrm{J})}\\&W=\frac{200000}{(26 \mathrm{J})}\end{aligned}\end$

W = 7692 .3 J.

Answer 2

${\bold{\huge{\red{\underline{\green{ANSWER}}}}}}$

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