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An ice cube with a mass of exactly 0.2843267 kg is sliding down an incline with

an acceleration of 5 m/s^2. Calculate the angle of the incline (in degrees).

I’m confused on how to find the angle can some explain?

User MBach
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1 Answer

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Answer:


30.7^(\circ)

Step-by-step explanation:

Assuming there is no friction, the only force acting on the ice cube along the incline is the component of the weight of the cube parallel to the incline, that is:


mg sin \theta

where

m is the mass of the cube

g is the acceleration of gravity


\theta is the angle of the incline

The equation of the forces along the incline therefore is


mg sin \theta = ma

where a is the acceleration.

Here we have the following data:


m=0.2843276 kg


g=9.8 m/s^2


a=5 m/s^2

Solving for
\theta, we find the angle of the incline:


\theta = sin^(-1) ((a)/(g))=sin^(-1)((5)/(9.8))=30.7^(\circ)

User AdamGold
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