Question 5The diagram given below shows a ski jump. A skier weighing 60kgf stands at A at the top of ski jump. He moves from A and takes off for his jump at B.(a) Calculate the change in the gravitational potential energy of the skier between A and B.(b) If 75% of the energy in part (a) becomes the kinetic energy at B, calculate the speed at which the skier arrives at B.(Take g = 10 ms-2).

Question

Accepted Answer

Given,Mass = 60 kg(a)Loss in potential energy=mg (h1 – h2)=60×10×(75−15)=60×10×60=3.6×104J\text {Loss in potential energy} = \text{mg (h1 – h2)} \[0.5em] = 60 \times 10\times (75-15) \[0.5em] = 60 \times10 \times 60 \[0.5em] = 3.6 \times 10^4 J \[0.5em]Loss in potential energy=mg (h1 – h2)=60×10×(75−15)=60×10×60=3.6×104J(b) When kinetic energy at B is 75% of (3.6 × 104)Kinetic energy at B=75100×3.6×104=27000J=2.7×104J\text{Kinetic energy at B} = \dfrac{75}{100} \times 3.6 \times 10^4 \[0.5em] = 27000J \[0.5em] = 2.7 \times 10^4 J \[0.5em]Kinetic energy at B=10075​×3.6×104=27000J=2.7×104JSince,Kinetic energy = 12\dfrac{1}{2}21​ mv2Substituting the values in equation we get,27000=12×60×v2⇒27000=1×30×v2⇒v2=2700030⇒v2=900⇒v=900⇒v=3027000 = \dfrac{1}{2} \times 60 \times v^2 \[0.5em] \Rightarrow 27000 = 1 \times 30 \times v^2 \[0.5em] \Rightarrow v^2 = \dfrac{27000}{30} \[0.5em] \Rightarrow v^2 = 900 \[0.5em] \Rightarrow v = \sqrt{900} \[0.5em] \Rightarrow v = 30 \[0.5em]27000=21​×60×v2⇒27000=1×30×v2⇒v2=3027000​⇒v2=900⇒v=900​⇒v=30∴ The speed at which the skier arrives at B = 30ms-1

Work, Energy and Power — Question 5

Question 5