Inclination of the plane .

An object of mass 1 kg is made to slide down a smooth inclined plane of length 20m. If the kinetic energy possessed by the body at the bottom of the plane is 100 J, what is the inclination of the plane with the horizontal (in degrees)?

Take g = 10 m / s 2 g = 10 m / s^2 .


The answer is 30.

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2 solutions

Lee Wall
Mar 29, 2014

By the principle of conservation of energy, the final kinetic energy is equal to the initial potential energy. The gravitational potential energy is given by U = m g h U = mgh , where m m is the object's mass, g g is the gravitational acceleration, and h h is the height of the object above the ground. Thus, we have 100 = 1 10 h 100 = 1 \cdot 10 \cdot h , which gives h = 10 h = 10 meters. Now let θ \theta be the inclination of the plane. By the Law of Sines, sin ( 90 ) 20 = sin θ 10 \frac{\sin(90)}{20} = \frac{\sin \theta}{10} . Therefore, θ = 30 \theta = \boxed{30} degrees.

Pradnya Patil
Oct 29, 2016

1/2mv^2=100 so v^2=200 but v^2=2as bcoz u=0, thus a=5m/s^2 but in sliding motion a=gain(angle of inclination)

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