Question

8.
A block of mass 'm' is resting on a rough horizontal plane of coefficient of friction is 'u', the minimum
work done in pushing the body through a distance 's' on the plane, when the force is applied at an angle
'O' with horizontal

1) umgs costheta/costheta+usintheta

2)umgs/sintheta+usintheta

3)umgscostheta/costheta-usintheta

4)umgs/costheta+usintheta

u stands for myu​

Answer 1

Explanation:

Correct option is

A

(mg+Qcosθ)

(P+Qsinθ)

f=P+Qsinθ, where f is the frictional force

μN=P+Qsinθ

μ(mg+Qcosθ)=P+Qsinθ

∴μ=

(mg+Qcosθ)

P+Qsinθ

Answer 2

Given:

Mass of block = m kg

Coefficient of friction of horizontal plane = μ

Distance through which block is pushed = s m

The angle at which force is applied = Ф

To Find:

Minimum work done in pushing the block through distance s

Solution:

• If we make the Free Body Diagram (FBD) of the given situation we get:

1. Force mg acting down on the block, balanced by Normal reaction applied by the plane.
2. Force F applied at angle Ф on the block is resolved as FcosФ horizontally and FsinФ downwards.
3. Force f (friction) acting opposite to FcosФ.

• Now we can find work done by resolving all the forces being applied on the block.

X-axis

                        ⇒ FcosФ - f = 0

                   Also,  f = μN

                  So, FcosФ - μN= 0                 (1)

Y-axis

                 ⇒  N - mg - FsinФ = 0

                 ⇒  N = mg + FsinФ                             (2)

• Substituting value of N from (2) in (1)

          ⇒ FcosФ - μ(mg + FsinФ) = 0

         ⇒ FcosФ = μmg + μFsinΦ

         ⇒ FcosФ - μFsinФ = μmg

         ⇒ F(cosФ - μsinΦ) = μmg

         ⇒ F = $\frac{umg}{cosO - usinO}$                                     (3)

• Now, work done is given by:

                W = F.s = FscosФ                         (dot product)

Substituting value from (3)

               W = $\frac{umgscosO}{cosO - usinO}$

Hence option 3) $\frac{umgscosO}{cosO - usinO}$ is the correct option.