Question

 

1) A meter stick is placed in a support stand and balanced at its center of gravity. A 50 gram mass is placed at the 25.00 cm mark. An unknown mass is placed on the other side of the meter stick and moved until the entire system is balanced. If the unknown mass is placed at the 75.00 cm mark, then what is the value of the unknown mass? Ignore the mass of the knife-edge clamps and assume that the meter stick is still balanced at its center of gravity. Show your work in the box below.

2) An experiment was performed using several masses on a spring. A graph of Force (N) vs. Elongation (m) was created and the equation of the line is: y = 11.0x + 0.107 From this information, what is the spring constant of the spring. Explain, in your OWN WORDS, how you determined this value.

3)In the thermal expansion lab, steam is passed through a metal rod. You measure the expansion of the rod to be 0.72 mm. The original length of the rod is 700.0 mm. The rod is heated from 25.0° C to 98.0° C. Calculate the coefficient of linear expansion of the rod.

4) A student performs an experiment to determine the coefficient of static friction between a wooden block and a wooden ramp. The ramp is slowly raised until the block just begins to slide and the angle of the ramp is measured. If the average angle that is measured is 40.0 degrees, then what is the coefficient of static friction? Show ALL WORK in the box below.

Answer 1

(1)

By taking moment of forces due to hanging mass about centre of gravity , we can get the unknown mass as follows. As seen from figure 50 gram mass is 25 cm left of centre of gravity and unknown mass is 25 cm right of centre of gravity

Counter clockwise moment = 50 10-3 g 25 N-m

Clockwise moment = m g 25 N - m

if we equate both the moments, we get unknown mass m = 50 gram

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(2)

when a spring is elongated by hanging different masses, we get the following releation

F = k ( x + xo ) ....................(1)

where F is applied force , k is spring constant, x is elongation due to applied force F ( weight of hanging mass ) and xo is the initial unstretched length of spring.

We have given equation , y = 11.0 x + 0.107 ................(2)

if we consider applied force F due to weight of hanging mass is plotted as a function of elongation distance, then by comparing eqn.(1) and (2), we get Force constant k = 11.0 N m-1

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(3)

If lo is the initial lenegth and l is length after thermal expansion, then we have

l = lo ( 1 + )

where is linear expansion coefficient, is temperature difference

expanded length = lo = 0.72 mm = 7.2 10-4 m

Hence, we get linear expansion coefficient as

= 0.7 / ( lo ) = ( 7.2 10-4 ) / ( 0.7 73 ) = 1.409 10-5 / oC

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(4)

When a block of mass m is on an inclined surface that makes angle with inclined surface, we get a component (mg sin ) of weight of block in the direction parallel to inclined surface and due to other component (mg cos) , we get equal normal reaction force from inclined surface to block of mass.

Hece static friction force that acts against the motion of block is mg cos.

where is static friction coefficient

If the angle of inclined surface gradually increases from low value, at particular angle the block start to slide down. At that instant friction force just balances the pulling force of block of mass

Hence we have, mg cos = mg sin   or = tan = tan40 = 0.84