Question

An object is released from rest at top at Plane at an the top of a rough plane at an angle of 30° to the horizontal. and 4.0m high. It the coettiment of friction is 0.3, calculate- take the time the body takes to reach the bottom of the plane Take [g=10m/s]​

Answer 1

It takes the object 2.581 seconds to reach the bottom of the plane.

Step-by-step explanation:

We will need to use Newton's Second Law to answer this question.

According to Newton's Second Law, acceleration is directly proportional to the net force and inversely proportional to the object's mass. Therefore the acceleration of an object depends on the mass of the object and the amount of force applied.

`F_(net)=ma`

Lets identify all the forces affecting the motion of the object.

The force due to gravity has two components; the horizontal (parallel) and vertical (perpendicular).

The force parallel to the plane trying to accelerate the block down the plane is

`mg\sin \theta`

The perpendicular force balances out with the normal force. The normal force is equal to the force of the object perpendicular to the plane.

`mg\cos\theta`

We also have a force opposing the motion of the object; the force due to friction. The formula for the force due to friction is

`F_F=F_N\mu`

`F_F=\mu mg\cos \theta`

We can say the net force of the object is

`ma=mg\sin \theta -\mu mg\cos \theta`

We can simplify this by cancelling the common factor of
`m`.

`a=g\sin \theta - \mu g\cos \theta`

We are given

`\theta=30 \textdegree\\\mu=0.3\\h=4\\g=10`

Substituting our values into the equation gives us

`a=10*\sin 30- 0.3* 10*\cos 30`

Acceleration

Lets simplify.

`a=10*(1)/(2) - 0.3* 10*\cos 30`

`a=10*(1)/(2) - 0.3* 10*(√(3) )/(2)`

`a=5- 0.3* 10*(√(3) )/(2)`

`a=5-3*(√(3) )/(2)`

`a=5-(3√(3) )/(2)`

`a=2.402`

Plane Length

We are given the angle and the side opposite to the angle.

We can use the SIN function to evaluate the length of the plane.

`O` is the opposite side (height)

`H` is the hypotenuse (plane length)

`\sin \theta=(O)/(H)`

Solving for
`H` gives us

`H=(O)/(\sin \theta)`

Given

`O=4\\\theta=30`

`H=(4)/(\sin 30)`

`H=(4)/((1)/(2) )`

`H=4*2`

`H=8`

The length of the plane is 8 meters.

Time

We can evaluate the time using this motion equation.

`\Delta x=v_ot+(1)/(2) at^2`

We are given

`\Delta x=8\\v_o=0\\a=2.402`

Substituting our values into the equation gives us

`8=0*t+(1)/(2) *2.402*t^2`

Lets simplify and evaluate
`t`.

`8=(1)/(2) *2.402*t^2`

`8=1.201*t^2`

`(8)/(1.201)=t^2`

`6.66=t^2`

`√(6.66)=t`

`2.581=t`