Question

A 60kg water-skier is accelerated from rest by a powerful boat through a tow rope that initially makes an angle of 10° down from the horizontal. It the rope provides a force of 1015 N along its length, what is the horizontal force acting on the skier? If the skier experiences a frictional drag due to the water of 400 N, what is her acceleration?

Answer 1

How do I draw the vector diagram for this question?

Answer 2

The force acting on the skier will be the horizontal component of the force = 1015 cos(10) = 999.6 N = 1000 N approx.
The acceleration = Net force /mass

Net force = 1000-400 N = 600N
Therfore acceleration = 600 /60 = 10 m/sec^2.

Answer 3

*Therefore

Answer 4

Thanks a lot!

Answer 5

You're welcome.