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noname [10]
3 years ago
10

Richardson pulls a toy 3.0 m across the floor by a string, applying a force of0.50 N. During the first meter, the string is para

llel to the floor. In the next two meters, the string makes an angle 0f 300 with the horizontal direction. What’s the total amount of work done by Richardson on the toy?
Physics
1 answer:
Anastasy [175]3 years ago
7 0

Answer:

Total Work done =0.65 joule

Explanation:

Work done is given Mathematically as

W=F *d

Where w=work done in joules

F=applied force

d= distance moved

The work done to move the toy accros the first meter is

W1=0.5*1

W1=0.5joule

The work done to move the toy across the next 2m at an angle of 30° is

.W2=0.5*2cos30

W2=0.5*2*0.154

W2=0.154joule

Hence total work done is

W1+W2=0.5+0.154

Total Work done =0.65 joule

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∑Fx' = m*aCM   ⇒    m*g*Sin θ - Ff = m*aCM

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m*g*Sin θ - Ff = m*g*Sin θ - (I*α / R) = m*aCM

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we get

aCM = (2/3)*g*Sin θ

6 0
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