Question

A body placed on an inclined plane just starts sliding downwards when the plane is inclined at an angle of 30°. Find the acceleration of the body (in m/s² to the nearest integer) when the plane is inclined at an angle of 45° with the horizontal plane. (assume uk = us and g = 10 m/s) (in m/s)​

Answer 1

anser is 5 under root3+E/200

Answer 2

Given: A body placed on inclined plane starts sliding when it is inclined at 30°

To find: Acceleration when plane inclined at 45°

Explanation: When the body just starts sliding, that means that the frictional force on the body is equal to the force provided by the horizontal component of force.

So, umg cos 30° = mg sin 30°

$u \times \frac{1}{2} = \frac{ \sqrt{3} }{2}$

=>$u = \sqrt{3}$

Let acceleration when angle= 45° be a.

Now, when angle is equal to 45°:

a= ug cos 45° - g sin 45°

=$\sqrt{3} \times 10 \times \frac{1}{ \sqrt{2} } - 10 \times \frac{1}{ \sqrt{2} }$

=$\frac{10}{ \sqrt{2} } ( \sqrt{3} - 1)$

=$5 \sqrt{2} ( \sqrt{3 } - 1)$

= 5*1.41 (1.73-1)

= 5*1.41*0.73

= 5.15 m/s^2

Therefore, the acceleration of the body when the plane is inclined at angle 45° is 5.15 m/s^2.