A 55 g golf ball is pushed up a 35° ramp. The ball starts from rest and the leng

Question

A 55 g golf ball is pushed up a 35° ramp. The ball starts from rest and the length of the surface of the ramp is 10 m. The coefficient of kinetic friction between the surface on the ramp and the ball is 0.2. The force stops when the ball leaves the top of the ramp and the total horizontal distance travelled by the ball after leaving the ramp is 30 m. What was the magnitude of the pushing force (in N)? The picture is of a ramp with an arc starting at the top and ending on the ground 30 m away. What is the best way at tackling this? Using something like kinematics or conservation of energy.

Explanation / Answer

I would approach it from a conservation of energy perspective. In this problem, there are four energy components to consider:
1. Potential energy of the ball at the top of the ramp (=mgh, where h is ramp length * sin(35) ),
2. Kinetic energy required to move the ball the 30m in the time it takes the ball to fall: solve for t in the formula 0 = (g/2)t2 + [V0*sin(35)]t + h, where V0 is the ball's velocity at the top of the ramp (unknown, leave t as a function of V0), then plug the t equation into V0*cos(35) = 30/t, and solve for V0. Now you can plug V0 into KE equation = 0.5*m*V02 .
3. Friction loss of the ball moving up the ramp (Fk = *Normal, Normal = mgsin(35) ). Remember Work = Force x distance, so Ek = Fk*10

4. Applied Force, FA.

You can either plug these into an explicit conservation equation, or just reason that FA must conter all the others, so FA*10 = KE + PE + EK . Solve for FK.

Hope this helps.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.