A kangaroo has a maximum gravitational potential energy during one jump of 770 J
1 answer:
Answer:
k = 44000 N/m
Explanation:
Given the following data;
Maximum gravitational potential energy = 770 J
Elastic potential energy = 14% of 770 J = 14/100 * 770 = 107.8 J
Extension, x = 42 - 35 = 7cm to meters = 7/100 = 0.07 m
To find the spring constant, k;
The elastic potential energy of an object is given by the formula;
Substituting into the equation, we have;
Cross-multiplying, we have;
k = 44000 N/m
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Answer:
A) The resultant force is 43.4 [N]
B) The movement of the heavy crate is going to the right and in the negative direction on the y-axis
Explanation:
We need to make a sketch of the different forces acting on the heavy crate.
In the attached image we can see the forces and the sum of the vector with their respective angles.
Forces in the X-axis
Forces in the y-axis
Using the Pythagorean theorem
The movement of the heavy crate is going to the right and in the negative direction on the y-axis, this can be easily seen in the graphical sum of vectors.
