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vodka [1.7K]
3 years ago
5

A bucket is filled partly with water such that its combined mass is 2.5 kg. It is tied to a rope and whirled in a circle with a

radius of 1.4 m. The speed at the top of the circle is 4.0 m/s and the speed at the bottom of the circle is 6.0 m/s. Determine the magnitude of the net force acting on the bucket at the bottom of the circle.
Physics
1 answer:
Ivenika [448]3 years ago
7 0

Answer:

1. Simply τ = m x g x r = 54kg x 9.8m/s² x 0.050m = 26 N·m

2. The bucket creates a torque

τ = 75kg x 9.8m/s² x 0.075m = 55 N·m,

so we must create the same torque with the handle.

55 N·m = F x 0.25m

F = 220 N

Explanation:

Hope this is helpful

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Answer:

The answer is "151.25 J and -547.64 J".

Explanation:

u = 0\\\\v = 2.35\  \frac{m}{sec}\\\\d = 5.0 \ m\\\\

Using formula:

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F_{net} = m \times a\\\\F_{net} = 55 \times 0.55 = 30.25\ N\\\\

Calculating the Work by net force

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Now,

F_{net} = F_p - F_f\\\\F_p = 140 \ N\\\\F_f = u_k\times m \times g = u_k \times 55 \times 9.81\\\\F_f = 539.55 \times u_k\\\\30.25 = 140 - u_k \times 55 \times 9.81\\\\u_k = \frac{(140 - 30.25)}{(55\times 9.81)}\\\\uk = 0.203 = \text{Coefficient of friction}\\\\W_f = -F_f \times d\\\\W_f = -0.203 \times 55 \times 9.81 \times 5\\\\Work\ done\  by\ friction = -547.64 \ J

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