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Physics Case study (4 Marks)

Work and Energy · UP Board (UPMSP) STD 11 Science (UPMSP - English Medium). 4 questions.

Questions

Case study (4 Marks)

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4 questions · self-marked practice — reveal the answer and mark yourself.

Question 14 Marks
In one of the exercises to strengthen the wrist and fingers, a person squeezes and releases a soft rubber ball. Is the work done on the ball positive, negative or zero during compression? During expansion?
Answer
  1. During compression, the work done on the ball is positive as the direction of the force applied by the fingers is along the compression of the ball.
  2. During expansion, the work done is negative as expansion takes place against the force applied by the fingers on the ball.
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Question 24 Marks
When you push your bicycle up on an incline the potential energy of the bicyle and yourself increases. Where does this energy come from?
Answer
When a person pushes his bicycle up on an inclined plane, the potential energies of the bicycle and the person increase because moving up on the inclined plane the kinetic energy decreases. and as mechanical energy is sum of kinetic energy and potential energy, and remains constant for a conservative system. Therefore, potential energy must increase in this case.
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Question 34 Marks
The US athlete Florence Griffith-Joyner won the $100m$ sprint gold medal at Seol Olympic $1988$ setting a new Olympic record of $10.54s$. Assume that she achieved her maximum speed in a very short-time and then ran the race with that speed till she crossed the line. Take her mass to be $50kg$.
  1. Calculate the kinetic energy of Griffith-Joyner at her full speed.
  2. Assuming that the track, the wind etc. offered an average resistance of one tenth of her weight, calculate the work done by the resistance during the run.
  3. What power GriffithJoyner had to exert to maintain uniform speed?
Answer
S = 100m, t = 10.54sec, m = 50kg The motion can be assumed to be uniform because the time taken for acceleration is minimum. Speed $\text{v}=\frac{\text{s}}{\text{t}}=9.487\text{m}/\text{s}$ So, K.E. $=\frac{1}{2}\text{mv}^2=2250\text{J}$ Weight = mg = 490J given $\text{R}=\frac{\text{mg}}{10}=49\text{J}$ so, work done against resistance $W_F = -RS = -49 × 100 = -4900J$ To maintain her uniform speed, she has to exert 4900J of energy to over come friction,$\text{P}=\frac{\text{W}}{\text{t}}$
$=\frac{4900}{10.54}=465\text{W}$
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Question 44 Marks
In tug of war, the team that exerts a larger tangential force on the ground wins. Consider the period in which a team is dragging the opposite team by applying a larger tangential force on the ground. List which of the following works are positive, which are negative and which are zero?
  1. Work by the winning team on the losing team.
  2. Work by the losing team on the winning team.
  3. Work by the ground on the winning team.
  4. Work by the ground on the losing team.
  5. Total external work on the two teams.
Answer
  1. Work by the winning team on the losing team is positive, as the displacement of the losing team is along the force applied by the winning team.
  2. Work by the losing team on the winning team is negative, as the displacement of the winning team is opposite to the force applied by losing team.
  3. Work by the ground on the winning team is positive.
  4. Work by the ground on the losing team is negative.
  5. Total external work on the two teams is positive.
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