Question: 19 A certain machine can outpur 50 Joules (J) of energy are input. If the amount of friction experienced by the madise is reduced, which of the following is most likely its outpun A. 0.3 B. 25.J C. 50.3 D. 75J I sim along a cafetert table with o constant ferce cf tit N. a food work do and do? 27J B 28J C 277J D. 18. Question: 20

Question 19:<br />A certain machine can output 50 Joules (J) of energy when 100 Joules (J) of energy are input. If the amount of friction experienced by the machine is reduced, which of the following is most likely its output?<br /><br />A. 0.3 J<br />B. 25 J<br />C. 50.3 J<br />D. 75 J<br /><br />Answer: D. 75 J<br /><br />Explanation: When the amount of friction experienced by the machine is reduced, the energy loss due to friction will decrease. Since the machine initially outputs 50 J of energy when 100 J are input, reducing the friction will result in a higher output. Therefore, the most likely output when the friction is reduced would be 75 J.<br /><br />Question 20:<br />A block of mass 2 kg is pushed along a carpeted table with a constant force of 10 N. How much work is done and what is the final velocity of the block?<br /><br />A. 27 J, 3 m/s<br />B. 28 J, 2.8 m/s<br />C. 27.7 J, 2.7 m/s<br />D. 18 J, 1.8 m/s<br /><br />Answer: B. 28 J, 2.8 m/s<br /><br />Explanation: The work done can be calculated using the formula Work = Force x Distance. The force applied is 10 N and the distance is 2.8 m, so the work done is 10 N x 2.8 m = 28 J.<br /><br />The final velocity of the block can be calculated using the formula Work = 1/2 x Mass x Velocity^2. Rearranging the formula to solve for velocity gives Velocity = sqrt(2 x Work / Mass). Substituting the values gives Velocity = sqrt(2 x 28 J / 2 kg) = 2.8 m/s.<br /><br />Therefore, the work done is 28 J and the final velocity of the block is 2.8 m/s.