4. An experiment was performed to determine the value of the gravitational accel

Question

4. An experiment was performed to determine the value of the gravitational acceleration equal g on Earth. Two masses of mass M hang at rest from the ends of a string on each side of a light pulley (see figure below). A mass m = 001M is placed on the right-hand-side mass. After the heavier side moved down from rest by h = im, has the small mass m is removed. The system continues to move for the next is, covering a distance of H = 0.312m. Find the value of 9. . . Up Figure 6: Pulley with a system of masses attached. (10 marks)

Explanation / Answer

Let tension be T and acceleration be a,

( M+m) a = Mg+mg - T

Ma = T - Mg

Adding both equation, a = mg/(2M+m)

= 0.01*g/(2+0.01) = 0.01/2.01 g

Using third equation of motion, v = sqrt (2 as)

= sqrt (2*a*1) = sqrt(2 a)

Now when mass m will be removed acceleration will be zero.

vt = H

sqrt(2a) *1 =0.312

2a = 0.312^2

2*0.01/2.01 g = 0.312^2

g = 0.312^2 ×2.01/0.02

= 9.783 m/s^2 Answer

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