Question

a solid ball is at the top of a 2 m tall ramp. if it is given a push so that it’s initial speed is 1.75 m/s, what is it’s speed when it reaches the bottom of the ramp?

Answer 1

6.50m/s

Step-by-step explanation:

Using conservation of energy, the total energy in the system is conserved. We know that the ball has potential energy initially because it is elevated 2 meters. We also know it has kinetic energy because it is given an initial speed of 1.75 m/s. At the bottom of the ramp, it loses all potential energy and is converted directly to kinetic energy. We can model the equation as below. U is the potential energy, K is the initial kinetic energy and K' is final kinetic energy.

`U + K = K'`

Equation for kinetic energy is

`K = 0.5mv^2`

and potential energy is:

`U=mgh`

Plug these equation to the first equation and simplify (v' is final speed):

`mgh+0.5mv^2= 0.5mv'^2\\gh+0.5v^2=0.5v'^2\\2gh+v^2=v'^2\\√(2gh+v^2)=v'`

Plug in the values:

`v'= √(2gh+v^2)=√(2(9.81)(2)+(1.75)^2) \\ v' = 6.50m/s`

Final speed is 6.50m/s