Discover a world of knowledge at Westonci.ca, where experts and enthusiasts come together to answer your questions. Join our platform to connect with experts ready to provide accurate answers to your questions in various fields. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform.

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 :

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