To solve this problem we will apply the concept of frequency in a string from the nodes, the tension, the linear density and the length of the string, that is,
Here
n = Number of node
T = Tension
= Linear density
L = Length
Replacing the values in the frequency and value of n is one for fundamental overtone
Similarly plug in 2 for n for first overtone and determine the value of frequency
Similarly plug in 3 for n for first overtone and determine the value of frequency
Answer:
The work done against gravity is 78.4 J
Explanation:
The work is calculated by multiplying the force by the distance that the
object moves
W = F × d, where W is the work , F is the force and d is the distance
The SI unit of work is the joule (J)
We need to find the work done against gravity when lowering a
16 kg box 0.50 m
→ F = mg
→ m = 16 kg, and g = 9.8 m/s²
Substitute these value in the rule
→ F = 16 × 9.8 = 156.8 N
→ W = F × d
→ F = 156.8 N and d = 0.50
Substitute these values in the rule
→ W = 78.4 J
<em>The work done against gravity is 78.4 J</em>
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