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julsineya [31]
2 years ago
8

In the lab setup above, block 1 is being pulled across a smooth surface by a light string connected across a pulley to a falling

block 2. The pulley has negligible mass and friction. Students measure the mass of each block, the distance from A to B , and the speeds of the block on the surface at points A and B . Which of the following hypotheses can be tested with this setupAs block 1 moves from point A to point B, the work done by the tension on block 1 is equal to the kinetic energy of block 1 at point B.As block 1 moves from point A to point B, the work done by the tension on block 1 is equal to the kinetic energy of the two-block system when block 1 is at point B.As block 1 moves from point A to point B , the work done by gravity on block 2 is equal to the kinetic energy of block 1 at point B.As block 1 moves from point A to point B, the work done by gravity on block 2 is equal to the change in the kinetic energy of the two-block system.As block 1 moves from point A to point B, the work done by the tension on block 2 is equal to the kinetic energy of block 2 when block I has reached point B.
Physics
1 answer:
Nuetrik [128]2 years ago
8 0

Answer:

As block 1 moves from point A to point B, the work done by gravity on block 2 is equal to the change in the kinetic energy of the two-block system.

Explanation:

As block 2 goes down , work is done by gravity on block 2 . This is converted

into kinetic energy of block 1 and block 2 . Work done by gravity is mgh which can be measured easily . kinetic energy of both the blocks can also be measured.  

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A Porsche 944 Turbo has a rated engine power of 217hp . 30% of the power is lost in the drive train, and 70% reaches the wheels.
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Explanation:

(a)  It is given that two-third of weight is over the drive wheels. So, mathematically, w = \frac{2}{3}mg.

Hence, maximum force is expressed as follows.

                F_{max} = \mu_{s} \times w

           m \times a_{max} = \mu_{s} (\frac{2}{3} mg)

Hence, the maximum acceleration is calculated as follows.

             a_{max} = \frac{2}{3} \mu_{s} \times g

                          = \frac{2}{3} \times 1.00 \times 9.8 m/s^{2}

                          = 6.53 m/s^{2}

Hence, the maximum acceleration of the Porsche on a concrete surface where μs = 1 is 6.53 m/s^{2}.

(b)  Since, 30% of the power is lost in the drive train. So, the new power is 70% of P_{max}.

That is,   new power = 0.7 \times P_{max}

Now, the expression for power in terms of force and velocity is as follows.

                      P = F_{max} \nu

              0.7 P_{max} = ma_{max} \nu

Therefore, speed of the Porsche at maximum power output is as follows.

            \nu = 0.7 \times \frac{P_{max}}{ma_{max}}

                      = 0.7 \times \frac{217 hp \times \frac{746 W}{1 hp}}{1500 kg \times 6.53 m/s^{2}}

                      = 11.568 m/s

                      = 11.57 m/s

Therefore, speed of the Porsche at maximum power output is 11.57 m/s.

(c)   The time taken will be calculated as follows.

             time = \frac{\text{velocity}}{\text{acceleration}}

                     = \frac{11.57 m/s}{6.53 m/s^{2}}

                     = 1.77 s

Therefore, the Porsche takes 1.77 sec until it reaches the maximum power output.

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