A small body is projected up a rough(coefficent of friction=0.5)inclined plane,(inclined angle=60 degree)with a speed 10 metre per second. How far along the plane it moves up before coming to rest?? g=10 ms^-2

To determine how far the small body moves up the inclined plane before coming to rest, we need to analyze the forces acting on it and apply the principles of kinematics. Let's break this down step by step.

Understanding the Forces at Play

When the body is projected up the inclined plane, two main forces act against its motion:

• Gravitational Force: This force acts downwards and has a component acting parallel to the incline, which works to pull the body back down.
• Frictional Force: Since the incline is rough, friction opposes the motion of the body. The frictional force can be calculated using the coefficient of friction and the normal force.

Calculating the Forces

First, let's find the gravitational force acting on the body. The weight of the body (W) can be expressed as:

W = m * g

Where:

• m is the mass of the body (which we can keep as a variable since it will cancel out later).
• g is the acceleration due to gravity, given as 10 m/s².

The component of the gravitational force acting down the incline (F_gravity) can be calculated as:

F_gravity = W * sin(θ) = m * g * sin(60°)

Using sin(60°) = √3/2, we have:

F_gravity = m * 10 * (√3/2) = 5√3 * m

Frictional Force Calculation

The normal force (N) acting on the body is given by:

N = W * cos(θ) = m * g * cos(60°)

Since cos(60°) = 1/2, we find:

N = m * 10 * (1/2) = 5m

The frictional force (F_friction) can be calculated as:

F_friction = μ * N = 0.5 * 5m = 2.5m

Net Force Acting on the Body

The total force acting against the motion of the body as it moves up the incline is the sum of the gravitational force and the frictional force:

F_net = F_gravity + F_friction = 5√3 * m + 2.5m

Factoring out m, we get:

F_net = m(5√3 + 2.5)

Using Kinematics to Find Distance

Now, we can apply the work-energy principle or kinematic equations to find the distance (d) the body travels before coming to rest. The initial velocity (u) is 10 m/s, and the final velocity (v) is 0 m/s. Using the equation:

v² = u² + 2a * d

We need to find the acceleration (a), which is negative since it opposes the motion:

a = - (5√3 + 2.5) * g/m

Substituting g = 10 m/s², we have:

a = - (5√3 + 2.5) * 10/m

Now, substituting into the kinematic equation:

0 = (10)² + 2 * (- (5√3 + 2.5) * 10/m) * d

Rearranging gives:

d = (10)² / (2 * (5√3 + 2.5) * 10/m)

Which simplifies to:

d = 10 / (5√3 + 2.5)

Final Calculation

Now we can calculate the distance:

First, let's evaluate the denominator:

5√3 ≈ 8.66 (using √3 ≈ 1.732)

Thus, 5√3 + 2.5 ≈ 8.66 + 2.5 = 11.16

Now substituting back into the distance formula:

d ≈ 10 / 11.16 ≈ 0.896 m

Therefore, the small body moves approximately 0.896 meters up the inclined plane before coming to rest.