What minimum speed does a 150g puck need to make it to the top of a 3.6m -long, 23 degree frictionless ramp?

the original KE must equal the PE at the height of the ramp

1/2 m v^2= mg*3.6sin23

To determine the minimum speed needed for the puck to make it to the top of the frictionless ramp, we need to consider the conservation of mechanical energy.

The mechanical energy of an object can be divided into two components: kinetic energy (KE) and gravitational potential energy (PE). At the top of the ramp, the puck will have zero kinetic energy (since it is momentarily at rest) and maximum potential energy (since it is at the highest point).

The equation for mechanical energy conservation is given by:

Initial KE + Initial PE = Final KE + Final PE

Since the puck starts at the bottom of the ramp, the initial KE is given as:

KE_initial = (1/2) * m * v^2

where m is the mass of the puck (150g = 0.15kg) and v is its initial velocity.

As the ramp is frictionless, there is no work done against friction, so the total mechanical energy is conserved. Therefore, the mechanical energy at the top of the ramp consists only of gravitational potential energy, given as:

PE_final = m * g * h

where g is the acceleration due to gravity (9.8 m/s^2) and h is the vertical height the puck has climbed, which can be calculated as:

h = ramp_length * sin(ramp_angle)

Substituting the values into the equation, we have:

(1/2) * m * v^2 + 0 = 0 + m * g * h

Simplifying and rearranging the equation for v, we get:

v = sqrt(2 * g * h)

Plugging in the values for g (9.8 m/s^2) and h (ramp_length * sin(ramp_angle)), we can find the minimum speed required for the puck to make it to the top of the ramp.