Open Question
A 6750-kg helicopter accelerates upward at 0.80m/s² while lifting a 1080-kg frame at a construction site, Fig. 4–66.
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(c) What force does the cable exert on the helicopter?
6. Intro to Forces (Dynamics)
Vertical Forces & Acceleration
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A 3-kg bucket is being pulled upwards by a cord. The tension in the cord is 35 N. What is the acceleration of the bucket? (The mass of the cord is negligible, which means you can assume m_cord = 0.)
A
25 m/s2
B
1.9 m/s2
C
21 m/s2
D
8.4 m/s2
Verified step by step guidance1
Identify the forces acting on the bucket: the tension in the cord (T = 35 N) and the gravitational force (weight) acting downwards. The weight can be calculated using the formula: \( F_{gravity} = m \cdot g \), where \( m = 3 \text{ kg} \) and \( g = 9.8 \text{ m/s}^2 \).
Calculate the gravitational force: \( F_{gravity} = 3 \text{ kg} \times 9.8 \text{ m/s}^2 \).
Determine the net force acting on the bucket. Since the bucket is moving upwards, the net force \( F_{net} \) is the difference between the tension and the gravitational force: \( F_{net} = T - F_{gravity} \).
Use Newton's second law to find the acceleration of the bucket. Newton's second law states that \( F_{net} = m \cdot a \), where \( a \) is the acceleration. Rearrange this formula to solve for \( a \): \( a = \frac{F_{net}}{m} \).
Substitute the values for \( F_{net} \) and \( m \) into the equation to calculate the acceleration: \( a = \frac{T - F_{gravity}}{m} \).
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