A77.0 kg woman slides down a 42.6 m long waterslide inclined at 42.3. at the bottom, she is moving 20.3 m/s. how much work did friction do on the woman? (unit=j)

5.791244495 KNm

Explanation:

The height h is given by,

Potential energy, PE is given by

PE=mgh where m is mass of the woman, g is acceleration due to gravity whose value is taken as and h is already given hence substituting 77 Kg for m we obtain

PE=21.6567095 KNm

We also know that Kinetic energy is given by where v is the velocity and substituting v for 20.3 we obtain

KE=15.865465 KNm

Friction work is the difference between PE and KE hence

Friction work=21.6567095 KNm-15865.465 Nm=5.791244495 KNm

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s

the components of a vector are the parts of the vectors that are to each other.

b perpendicular