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lord [1]
2 years ago
11

A sled drops 50 meters in height on a hill. The mass of the rider and sled is 70 kg and the sled is going 10 m/s at the bottom o

f the hill. What is the efficiency of energy conversion from potential to kinetic?
Physics
1 answer:
navik [9.2K]2 years ago
4 0

Answer:

Efficiency = 10.2 %

Explanation:

Given the following data;

Mass = 70 kg

Height = 50 m

Velocity = 10 m/s

We know that acceleration due to gravity is equal to 9.8 m/s².

To find the efficiency of energy conversion from potential to kinetic;

First of all, we would determine the potential energy;

P.E = mgh

P.E = 70 * 9.8 * 50

P.E = 34300 J

For the kinetic energy;

K.E = ½mv²

K.E = ½ * 70 * 10²

K.E = 35 * 100

K.E = 3500

Therefore, Input energy, I = 34300 J

Output energy, O = 3500 J

Next, we find the efficiency;

Efficiency = O/I * 100

Substituting into the formula, we have;

Efficiency = 3500/34300 * 100

Efficiency = 0.1020 * 100

Efficiency = 10.2 %

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Robin would like to shoot an orange in a tree with his bow and arrow. The orange is hanging yf=5.00 myf=5.00 m above the ground.
tensa zangetsu [6.8K]

Answer:

h' = 55.3 m

Explanation:

First, we analyze the horizontal motion of the projectile, to find the time taken by the arrow to reach the orange. Since, air friction is negligible, therefore, the motion shall be uniform:

s = vt

where,

s = horizontal distance between arrow and orange = 60 m

v = initial horizontal speed of the arrow = v₀ Cos θ

θ = launch angle = 30°

v₀ = launch speed = 35 m/s

Therefore,

60 m = (35 m/s)Cos 30° t

t = 60 m/30.31 m/s

t = 1.98 s

Now, we analyze the vertical motion to find the height if arrow at this time. Using second equation of motion:

h = Vi t + (1/2)gt²

where,

Vi = Vertical Component of initial Velocity = v₀ Sin θ = (35 m/s)Sin 30°

Vi = 17.5 m/s

Therefore,

h = (17.5 m/s)(1.98 s) + (1/2)(9.81 m/s²)(1.98 s)²

h = 34.6 m + 19.2 m

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since, the arrow initially had a height of y = 1.5 m. Therefore, its final height will be:

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<u>h' = 55.3 m</u>

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2 years ago
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