Physics, asked by saryka, 2 months ago

Q1. A block of weight of 1N rests on an inclined plane of inclination θ with the horizontal. The coefficient of friction between the block and the inclined plane is μ. The minimum force that has to be applied parallel to the inclined plane to make the body just move up the plane is?​

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Answers

Answered by IdyllicAurora
69

Concept :-

Here the concept of Friction and Force has been used. We see that we are given the weight of the block which is resting on the inclined surface. Even we are given the coefficient of friction. So first we can find out the components of force acting on the block. Then we have to find the minimum required force which will given us the answer.

Let's do it !!

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Solution :-

Given,

» Angle of inclination of plane with horizontal = θ

» Weight of the block = W = 1 N

» Coeffcient of friction between the block and plane = μ

  • Let the force required to move the block just upward be F

We know that, Weight = Mass (m) × Acceleration

Here acceleration will be due to gravity that is g.

So,

Weight = m × g

In vector resolution, direction is also referred in the formula.

  • Weight is a type of force only.

Now let's understand the figure. The vertical component of force acting on the body is W sinθ. The horizontal component of force acting on the body will be W cos θ . Now in order to make the block in balance there is friction also. This friction is in the direction of horizontal component of force. So the frictional force will be μ W cos θ .

From here we get,

  • Gravitational Force = Force acting due to gravity downwards = W sinθ

  • Frictional Force = Force acting between the layer of block and inclined surface = μ W cos θ

Now we see that, we need the minimum force of moving the block just upward. So it must be equal to the sum of all opposing forces. So,

>> F = Frictional Force + Gravitational Force (Weight acting downwards

By applying values,

>> F = W sin θ + μW cos θ

>> F = (m•g)sin θ + μ(m•g)cos θ

We already know that m × g = 1

>> F = (1)sin θ + μ(1)cos θ

>> F = sin θ + μ cos θ

>> F = μ cos θ + sin θ

This is the required answer.

Hence, option D.) μ cos θ + sin θ is correct.

\;\underline{\boxed{\tt{Required\;\:Force\;=\;\bf{\purple{\mu\cos\theta\;+\;\sin\theta}}}}}

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More to know :-

Workdone = Force × Distance

Momentum = Mass × Velocity

Velocity = Displacement / Time

Power = Workdone / Time

Energy = Workdone

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