(1) 8408 J = 8.4 kJ
We need to find the acceleration of the spelunker first, through the equation:
![v^2-u^2=2ad](https://tex.z-dn.net/?f=v%5E2-u%5E2%3D2ad)
where
v = 4.40 m/s is the final speed
u = 0 is the initial speed
a is the acceleration
d = 10.0 m is the distance
Solving the equation for a, we find
![a=\frac{v^2-u^2}{2d}=\frac{(4.40 m/s)^2-0}{2(10.0 m)}=0.97 m/s^2](https://tex.z-dn.net/?f=a%3D%5Cfrac%7Bv%5E2-u%5E2%7D%7B2d%7D%3D%5Cfrac%7B%284.40%20m%2Fs%29%5E2-0%7D%7B2%2810.0%20m%29%7D%3D0.97%20m%2Fs%5E2)
In order to find the magnitude of force F used to lift the spelunker, we have to apply Newton's second law:
![\sum F = ma\\F - mg = ma](https://tex.z-dn.net/?f=%5Csum%20F%20%3D%20ma%5C%5CF%20-%20mg%20%3D%20ma)
where (mg) is the weight of the spelunker, and a = 0.97 m/s^2. Solving for F, we find
![F=m(g+a)=(78.0 kg)(9.81 m/s^2+0.97 m/s^2)=840.8 N](https://tex.z-dn.net/?f=F%3Dm%28g%2Ba%29%3D%2878.0%20kg%29%289.81%20m%2Fs%5E2%2B0.97%20m%2Fs%5E2%29%3D840.8%20N)
And so, the work done by the force during this stage is
![W=Fd=(840.8 N)(10.0 m)=8408 J](https://tex.z-dn.net/?f=W%3DFd%3D%28840.8%20N%29%2810.0%20m%29%3D8408%20J)
(2) 7644 J = 7.6 kJ
The work done on the spelunker in this stage is
![W=Fd](https://tex.z-dn.net/?f=W%3DFd)
where F is the force applied on the spelunker to lift him, and d = 10.0 m is the vertical distance through which the spelunker is lifted.
In order to find the magnitude of F, we have to apply Newton's second law:
![\sum F = ma\\F - mg = ma](https://tex.z-dn.net/?f=%5Csum%20F%20%3D%20ma%5C%5CF%20-%20mg%20%3D%20ma)
where (mg) is the weight of the spelunker, and the acceleration is zero because he is moving at constant speed: so, a=0, and the equation becomes
![F-mg=0\\F=mg=(78.0 kg)(9.8 m/s^2)=764.4 N](https://tex.z-dn.net/?f=F-mg%3D0%5C%5CF%3Dmg%3D%2878.0%20kg%29%289.8%20m%2Fs%5E2%29%3D764.4%20N)
So, the work done is
![W=(764.4 N)(10.0 m)=7644 J](https://tex.z-dn.net/?f=W%3D%28764.4%20N%29%2810.0%20m%29%3D7644%20J)
3) 6895 J = 6.9 kJ
This stage is similar to stage (1); we find the deceleration using:
![v^2-u^2=2ad](https://tex.z-dn.net/?f=v%5E2-u%5E2%3D2ad)
where
v = 0 m/s is the final speed
u = 4.40 is the initial speed
a is the acceleration
d = 10.0 m is the distance
Solving the equation for a, we find
![a=\frac{v^2-u^2}{2d}=\frac{0-(4.40 m/s)^2}{2(10.0 m)}=-0.97 m/s^2](https://tex.z-dn.net/?f=a%3D%5Cfrac%7Bv%5E2-u%5E2%7D%7B2d%7D%3D%5Cfrac%7B0-%284.40%20m%2Fs%29%5E2%7D%7B2%2810.0%20m%29%7D%3D-0.97%20m%2Fs%5E2)
In order to find the magnitude of force F used to lift the spelunker, we have to apply Newton's second law:
![\sum F = ma\\F - mg = ma](https://tex.z-dn.net/?f=%5Csum%20F%20%3D%20ma%5C%5CF%20-%20mg%20%3D%20ma)
where (mg) is the weight of the spelunker, and a = -0.97 m/s^2. Solving for F, we find
![F=m(g+a)=(78.0 kg)(9.81 m/s^2-0.97 m/s^2)=689.5 N](https://tex.z-dn.net/?f=F%3Dm%28g%2Ba%29%3D%2878.0%20kg%29%289.81%20m%2Fs%5E2-0.97%20m%2Fs%5E2%29%3D689.5%20N)
And so, the work done by the force during this stage is
![W=Fd=(689.5 N)(10.0 m)=6895 J](https://tex.z-dn.net/?f=W%3DFd%3D%28689.5%20N%29%2810.0%20m%29%3D6895%20J)