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A 74.2 kg teenager slides down a frictionless slide that has an angle of inclination of 38˚. What is the magnitude of the acceleration of the teenager down the slide?

Sagot :

Luv2Teach

Answer:

Explanation:

This is one-dimensional motion on an incline which makes it so much more funner :/

The equation for this is

w*sinθ - f = ma which is just a fancy version of Newton's second law that says that the sum of the forces acting on an object are equal to that object's mass times its acceleration.

w is the weight of the teen which is also the normal force, [tex]F_n[/tex]. Therefore,

[tex]w=F_n=mg[/tex] where m is mass and g is gravity:

[tex]w=F_n=(74.2)(9.8)[/tex] and we need 2 sig figs for this since there are 2 in 9.8:

w = 730N

Since there is no friction in this problem, the value for f, frictional force, is 0, making our job a bit shorter. Filling in the main equation:

730sin38 - 0 = 74.2a

Since the left side is subtraction and we will eventually have to divide, and the rules for subtraction/addition are different from those for multiplication/division, we will have to subtract first, get the correct number of sig figs for that difference, and then divide, following the rules for division.

730sin38 will have 2 sig figs which is 450. Now

450 - 0 = 74.2a and

450 = 74.2a and divide to 2 sig figs:

a = 6.1 m/s/s

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