A small child gives a plastic frog a big push at the bottom of a slippery 2.0 meter long, 1.0 meter high ramp, starting it with
a speed of 5.0 m/s. What is the frog's speed as it flies off the top of the ramp?
1 answer:
Refer to the diagram shown below.
Because the ramp is slippery, ignore dynamic friction.
Let m = the mass of the frog.
g = 9.8 m/s²
The KE (kinetic energy) at the bottom of the ramp is
KE₁ = (1/2)*(m kg)*(5 m/s)² = 12.5 m J
Let v = the velocity at the top of the ramp.
The KE at the top of the ramp is
KE₂ = (1/2)*m*v²= 0.5 mv² J
The PE (potential energy) at the top of the ramp relative to the bottom is
PE₂ = (m kg)*(9.8 m/s²)*(1 m) = 9.8m J
Conservation of energy requires that
KE₁ = KE₂ + PE₂
12.5m = 0.5mv² + 9.8m
0.5v² = 2.7
v = 2.324 m/s
Answer: 2.324 m/s
Because the ramp is slippery, ignore dynamic friction.
Let m = the mass of the frog.
g = 9.8 m/s²
The KE (kinetic energy) at the bottom of the ramp is
KE₁ = (1/2)*(m kg)*(5 m/s)² = 12.5 m J
Let v = the velocity at the top of the ramp.
The KE at the top of the ramp is
KE₂ = (1/2)*m*v²= 0.5 mv² J
The PE (potential energy) at the top of the ramp relative to the bottom is
PE₂ = (m kg)*(9.8 m/s²)*(1 m) = 9.8m J
Conservation of energy requires that
KE₁ = KE₂ + PE₂
12.5m = 0.5mv² + 9.8m
0.5v² = 2.7
v = 2.324 m/s
Answer: 2.324 m/s
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Answer:
(a)Therefore the highest altitude attained by the object is =576 ft .
(b)Therefore the object takes 6 sec to fall to the ground.
Explanation:
Initial velocity: Initial velocity is a velocity from which an object starts to move.
u is usually used for notation of initial notation.
Final velocity: Final velocity is a velocity of an object after certain second from starting.
The final velocity is denoted by v.
Acceleration: The difference of final velocity and initial velocity per unit time
The S.I unit of acceleration is m/s².
(a)
Given that u= 128 ft\sec and g = 32 ft/sec².
At highest point the velocity of the object is 0 i.e v=0
Since the displacement is opposite to the gravity.
Therefore acceleration( a)= -g = -32 ft/sec².
To find the time this to happen we use the following formula
Here v=0
⇒0=128+(-32) t
⇒32t=128
⇒t = 4 sec
To determine the height we use the following formula
⇒s= 256 ft
Therefore the highest altitude attained by the object is =(320+256)ft=576 ft .
(b)
At the highest point the velocity of the object is 0.
so u=0. a=g= 32 ft/sec² [ since the direction of gravity and the displacement are same] s= 576 ft
To determine the time to fall we use the following formula
⇒t=6 sec
Therefore the object takes 6 sec to fall to the ground.