The diagram depicts a system at $t = 0$ . The work done by tension in the string, from $t = 0$ to $t = 1$ is $50 α$ Joules. The value of $α$ is

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Solution

Total acceleration of system

f = ( $m_{1} + m_{2})$ a

a = $\frac{f}{m_{1} + m_{2}}$

= $\frac{40}{2} = 20 m / s^{2}$

form t = 0 to t = 1

v = u + at

= 0 + 20 ( 1 )

v = 20

so from work energy theorem

w = $Δ$ ke

$50 α$ = $\frac{1}{2} m v^{2} - \frac{1}{2} m u^{2}$

$50 α = \frac{1}{2} (1) \times 400$

$α$ = 4

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The diagram depicts a system at t=0. The work done by tension in the string, from t=0 to t=1 is 50α Joules. The value of α is

Total acceleration of system f = ( m1+m2) a a = fm1+m2 = 402=20m/s2form t = 0 to t = 1v = u + at = 0 + 20 ( 1 ) v = 20 so from work energy theoremw = Δ ke50α = 12mv2−12mu250α=12(1)×400α = 4

The diagram depicts a system at $t = 0$ . The work done by tension in the string, from $t = 0$ to $t = 1$ is $50 α$ Joules. The value of $α$ is