Question

1) A 68 kg skydiver falls at a contast terminal speed of 59m/s. at what rate is gravitational potential energy being removed from the earth-skydiver system in kw?

2) A 2.0 kg block is dropped from a height of 40cm onto a spring of force constant k= 1960 n/m. what is the max displacement of the spring in cm?

Answer 1

1). Inital Speed of skydiver, u= 0m/s (Since he falls from the height thus initially was at rest)

Final speed, v= 59 m/s

Acceleration acting on him, a= +g= +9.8 m/s² (Since gravity supports the motion)

then let time taken be "t" and he falls fom the height "h". Then using equations of motion,

  

  

  .........................(1)

and

..........................(2)

then rate of loss of gravitational potential energy is,

using equation 1 and 2 in above,

(ANS)

2). Mass of block, m= 2.0kg

Height of block, h= 40cm=0.4m

Spring constant, k= 1960 N/m

Let maximum displacement is "x_m" then Using Energy Conservation Principle,

Gravitational Potential ENergy= Maximum Spring Potential Energy

  

Using given values in above,

  (ANS)