Can someone explain a couple physics problems involving Newton's laws
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Can someone explain a couple physics problems involving Newton's laws

[From: ] [author: ] [Date: 12-03-16] [Hit: ]
A 44.1 N weight is hung on a spring scale, and the scale is hung on a string. The string is lowered at a rate such that the entire assembly has a downward acceleration of 3m/s^2. How much force does the scale read?A car moves around a circular track.......
Hi, I'm really horrible at physics and have a test tomorrow.

Can anyone explain these questions to me? Please detail the equations and how you get to the answer, because I do want to understand it. Thanks so much.

A 44.1 N weight is hung on a spring scale, and the scale is hung on a string. The string is lowered at a rate such that the entire assembly has a downward acceleration of 3m/s^2. How much force does the scale read?

A car moves around a circular track. It is 28.2m from the center of the track and accelerates at 8.05m/s^2. What with what tangential velocity does it move? What is giving the car its centripedal force?

Determine the stopping distance for a skier moving down a slope with an initial speed of 20m/s and an angle of 5 degrees. Assume the slope is rough and has a coefficient of kinetic friction of .18.

(see picture below, figure on left)

A conical pendulum is a bob moving in a horizontal circle at the end of a long wire. The angle between the wire and the vertical does not change. A conical pendulum with an 80kg mass on a 10m wire swings at a 5 degree angle with the vertical. For the pendulum, determine (a) the tension in the wire and (b) the centripedal acceleration that the mass experiences.
(see picture below, figure on right)

Picture: http://i41.tinypic.com/2ll2fb5.png

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1. Net force acting on the weight = [(44.1/9.81)*3] N downwards
Force exerted by earth on the weight = 44.1 N

So spring must apply a force "F" N upwards so that
44.1 - F = [(44.1/9.81)*3] or
F = 44.1 - 13.5 = 30.6 N
This reading will be shown by the scale

2. tangential velocity = sq rt[8.05*28.2] = 15.066 or 15.1 m/s
Spring is extended to provide the centripetal force to the car.

3.Stopping distance, s = (20^2)/[2*{0.18*mg*cos 5 - mg*sin 5}/m] or
s = 200/[(9.81)*{0.18*cos 5 - sin 5}] = 221.2 m or 221 m

4.Let centripetal acceleration be "a" m/s^2 and "T be the tension of the string. We have
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