Question

A block of mass m slides down from rest from the top of a rough inclined plane at 45 degree with u = 0.5. Find the speed of particle when it reaches at the bottom​

Answer 1

Concept:

One dimensional motion

Motion on an inclined plane

Given:

The block starts from rest, so initial speed u = 0 m/s

The angle of incline θ = 45°

The coefficient of friction μ = 0.5

The mass of the block = m

Find:

The speed of the block when it reaches the bottom

Solution:

The frictional force acting on the block = μN

where N is the normal force

The normal force N = mg cos 45°

the frictional force = μ mg cos 45° = μmg cos 45°

The force pulling the block downwards = mg sin 45°

the net force acting on the block = mg sin 45° - μmg cos 45°

The net acceleration on the block = g sin 45° - μg cos 45°

The net acceleration on the block = 6.93 - 3.46 = 3.46 m/s^2

We know the kinematics equation

v² = u² + 2as

v² = 0 + 2 (3.46) s

v = 2.63√s

The speed of the block is 2.63√s m/s.

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Answer 2

Answer:

Explanation:

Given:

Initial speed u = 0 m/s (Block is at rest)

The angle of incline θ = 45°

The coefficient of friction μ = 0.5

The mass of the block = m

Solution:

The frictional force acting on the block = μN

where N is the normal force

The normal force N = mg cos 45°

the frictional force = μ mg cos 45° = μmg cos 45°

The force pulling the block downwards = mg sin 45°

the net force acting on the block = mg sin 45° - μmg cos 45°

The net acceleration on the block = g sin 45° - μg cos 45°

The net acceleration on the block = 6.93 - 3.46 = 3.46 m/s^2

Using the kinematics equations

$v^2+u^2=2as\\v^2=0+2*3.46\\v=2.63$

The speed of the block is 2.63√s m/s.

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