A sled and rider with a combined mass of 70 kg are at the top of a hill that rises 9 m above the level ground below. The sled is
given a push, providing an initial kinetic energy at the top of the hill of 1,200 J. a. Choosing a reference level at the bottom of the hill, what is the potential energy of the sled and rider at the top of the hill? sled and rider at the top of the hill? of the sled and rider at the bottom of the hill? b. After the push, what is the total mechanical energy of the hill?c. If friction can be ignored, what will be the kinetic energy of the sled and rider at the bottom of the hill?
1 answer:
Answer:
(A) 6174J, 0J.
(B) 7374J
(C) 7374J
Explanation:
See the attachment below for the calculation.
The princi6of conservation of energy has been used.
E = K1 + U1 = K2 + U2
Where E = total mechanical energy.
Taking the bottom as reference, and h =0 and as a result U2 = mg(0) = 0J.
The complete solution to the problem can be found in the attachment below.
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Answer:
The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.
Explanation:
Given that,
Mass = 2.15 kg
Distance = 0.0895 m
Amplitude = 0.0235 m
We need to calculate the spring constant
Using newton's second law
Where, f = restoring force
Put the value into the formula
We need to calculate the kinetic energy of the mass
Using formula of kinetic energy
Here,
Here,
Put the value into the formula
Hence, The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.