Acceleration of gravity

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An object is released from rest and falls in free fall motion. The speed v of the object after it has fallen a distance y is given by

v2 = 2gy.
In an experiment, v andy are measured and the measured values are used to calculate g. If the percent uncertainty in the measured value of v is 4.13% and the percent uncertainty in the measured value of y is 5.00%, what is the percent uncertainty in the calculated value of g? (Do not enter units for this answer.)
Oct 3rd, 2015

Thank you for the opportunity to help you with your question!

Given v^2=2gy   ....(1)

Differentiating this above equation with respect to v, we then have

2v=2g(dy/dv)+2y(dg/dv)

2vdv=2g(dy)+2y(dg)

Dividing both sides with v^2, we have

2(dv/v)=(2g/v^2)dy+(2y/v^2)dg

Grom equation (1), we have

2(dv/v)=(dy/y)+(dg/g)   ----(2)

Given that dy/y=0.05 and dv/v=0.0413

Substituting this in equation(2),  we have

2x0.0413=0.05+dg/g

dg/g=0.0326

(dg/g)x100=3.26%

the percent uncertainty in the calculated value of g is 3.26%

Please let me know if you need any clarification. I'm always happy to answer your questions.
Sep 21st, 2015

Thank you, Could you also help me with this one?

A small block travels up a frictionless incline that is at an angle of 30.0° above the horizontal. The block has speed 4.21 m/s at the bottom of the incline. Assume

g = 9.80 m/s2.
How far up the incline (measured parallel to the surface of the incline) does the block travel before it starts to slide back down?
Sep 21st, 2015

ok...just wait

Sep 21st, 2015

acceleration on the frictionless incline a==-gsin(30)=-g/2

initial velocty=u=4.21 m/s

Final velocity=v=0

Using kinematics equation, v^2=u^2-2aS

S is the distance travelled by the block.

as v=0, u^2=2aS

S=u^2/2a=4.21^2/2(g/2)=4.21^2/g=17.7241/9.8=1.8 metre

Sep 21st, 2015

Do contact me, if any further assistance is required...

Pallvee

Sep 21st, 2015

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Oct 3rd, 2015
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Oct 3rd, 2015
Nov 23rd, 2017
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