A roller-coaster car is moving at 20.0 m/s along a straight horizontal track. What is its speed after climbing a 15.0 m hill? Neglect the effects of friction.
Without friction and air resistance, we equate potential (PE) and kinetic (KE)energies.
Energies at bottom of hill = energies at top of hill.
0+(1/2)m*vi²=mgh+(1/2)m*vf^sup2;
cancel m and substitute
g=9.8 m/s²
vi=20 m/s
h=15 m
(1/2)(20²)=9.8*15+(1/2)vf²
Solve for vf.
HELP !! I do not fully understand how to completely solve this problem. Can you please help me further with the equation?? Thank You. !
5.14
10
To determine the speed of the roller-coaster car after climbing the hill, we can use the principle of conservation of energy.
The roller-coaster car initially has kinetic energy due to its motion along the track. This kinetic energy will be converted into potential energy as the car climbs the hill.
The potential energy gained by an object of mass m when it is lifted to height h can be calculated using the formula:
Potential energy (PE) = mass (m) × gravity (g) × height (h)
Since the car is neglecting any effects of friction, we can assume that the potential energy gained is equal to the kinetic energy lost.
The initial kinetic energy of the car can be calculated using the formula:
Kinetic energy (KE) = 0.5 × mass (m) × velocity (v)^2
We are given the initial velocity of the car (v = 20.0 m/s) and the height of the hill (h = 15.0 m).
First, calculate the initial kinetic energy of the car:
KE = 0.5 × m × v^2
Next, calculate the potential energy gained by the car:
PE = m × g × h
Since the potential energy gained is equal to the kinetic energy lost, we can equate the two equations:
KE = PE
0.5 × m × v^2 = m × g × h
Divide both sides of the equation by m:
0.5 × v^2 = g × h
Finally, solve for the final velocity (v'):
v' = sqrt(2 × g × h)
Substitute the given values:
v' = sqrt(2 × 9.8 m/s^2 × 15.0 m)
Calculate the final velocity to find the answer.