Assighment question based on Tension...

[Krist]

1. A waterskier of mass 80kg, starting from rest, is pulled along in a northerly direction by a horizontal rope with a constant tension of 240 Newtons. After 6 seconds he has reached a speed of 12m/s.

1. What is the net force on the skier?
2. If the tension in the rope were the only horizontal force acting on the skier, what would his accelleration be?
3. What is the sum of the resistance forces on the skier.

2. Any of newtons laws + constant acceleration



3. Ok guys, what i've managed to gather is the acceleration is 2m/s/s, initial velocity=0 and obviosly tenstion = 240N

To get the mass moving without resistance would require a force of 160N.
F=ma
F= 80 *2
F=160N

Now if i add them together will i get net force? NetF= 160 + 240 = 400N?
I'm not sure on that one, i am not clear about what to do with the tension measurement...does it affect both sides? Does it mean that the skier is only getting pulled with half the ammount of Force?
For Q2.
F = ma
240 = 80 * a
a = 3m/s/s

^ Is that right?
Finally for Q3, i can only guess taking 160 away from 240...bringing resistance to 80N.
I'm really not sure on any of these and our teacher kind of flew threw tension section.

[alphysicist]

Hi Krist,

A couple of your statements don't seem right to me.

1. A waterskier of mass 80kg, starting from rest, is pulled along in a northerly direction by a horizontal rope with a constant tension of 240 Newtons. After 6 seconds he has reached a speed of 12m/s.

1. What is the net force on the skier?
2. If the tension in the rope were the only horizontal force acting on the skier, what would his accelleration be?
3. What is the sum of the resistance forces on the skier.

2. Any of newtons laws + constant acceleration



3. Ok guys, what i've managed to gather is the acceleration is 2m/s/s, initial velocity=0 and obviosly tenstion = 240N

To get the mass moving without resistance would require a force of 160N.

That's true; but the point is even with resistance it still requires net force of 160N. In terms of the net force in the horizontal direction you have:


F_{\mbox{net},x} = m a_x



F=ma
F= 80 *2
F=160N

Now if i add them together will i get net force? NetF= 160 + 240 = 400N?

The 160N is the net force. Writing the equation in terms of the individual forces instead of the net force gives, for the horizontal direction:


\begin{align}
\sum F_x &= m a_x\nonumber\\
T_x + R_x &= m a_x\nonumber
\end{align}

which gives the answer you found in question 3.


I'm not sure on that one, i am not clear about what to do with the tension measurement...does it affect both sides? Does it mean that the skier is only getting pulled with half the ammount of Force?
For Q2.
F = ma
240 = 80 * a
a = 3m/s/s

^ Is that right?
Finally for Q3, i can only guess taking 160 away from 240...bringing resistance to 80N.
I'm really not sure on any of these and our teacher kind of flew threw tension section.