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belka [17]
3 years ago
5

A constant force of 12N is applied for 3.0s to a body initially at rest. The final velocity of the body is 6.0ms–1. What is the

mass of the body?
Physics
1 answer:
sp2606 [1]3 years ago
4 0
From the question,
u = 0m {s}^{ - 1}
v = 6m {s}^{ - 1}

t = 3s
F=12N



Using Impulse, the product of the constant force, F and time t equals the product of the mass of the body and change in velocity.

Ft =m(v-u)


12(3.0)=m(6.0- \: 0)
This implies that

36.0 = 6m
m =  \frac{36.0}{6.0}
\therefore \: m = 6.0kg


You can also use the equation of linear motion,
v = u + at
6 = 0 + a(3)
6 = 3a
a =  \frac{6}{3}

a = 2 {ms}^{ - 2}
But
F=ma
12 = m(2)
12 = 2m
\frac{12}{2}  = m
\therefore \: m = 6kg
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Explanation:

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Hence

     0.85 =  1 -  e^{-\frac{2.4}{\tau } }

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        \tau  = 1.265 s

     

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